CHE 116: General Chemistry
CHAPTER
FOURTEEN
Copyright © Tyna L. Heise 2002
All Rights Reserved
Chem 116, Prof. T.L. Heise
1
Chemical Kinetics
The area of chemistry that is concerned with
the speeds, or rates, at which reactions
occur.
The rates of chemical reactions are affected by
several factors, most notably:
Concentration of reactants
Temperature at which the reaction occurs
The presence of a catalyst
The surface area of solid or liquid
reactants or catalysts
Chem. 14.1
Chem 116, Prof. T.L. Heise
2
Chemical Kinetics
The speed of an event is defined as the change
that occurs in a given interval of time:
- commonly referred to as the
reaction rate
- reaction rate is a measure of how
quickly A is consumed or how quickly B is
produced
Average rate = change in # of moles of B
change in time
Chem. 14.1
Chem 116, Prof. T.L. Heise
3
Chemical Kinetics
Average rate = D(moles of B)
Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
4
Chemical Kinetics
The rate expression focuses on the number of
moles of b that are produced during the
reaction; it is the rate of appearance of B.
Chem. 14.1
Chem 116, Prof. T.L. Heise
5
Chemical Kinetics
The rate expression could just as easily been
described in the terms of the change in
moles of A, or the rate of disappearance of A.
-D(moles of A) = D(moles of B)
Chem. 14.1
Chem 116, Prof. T.L. Heise
6
Chemical Kinetics
For primarily all reactions, the reaction rates
will be determined by following changes in
concentration.
Units for reaction rates will be M/s.
- brackets around a chemical substance
indicate the concentration of the
substance.
- the minus sign always indicates the
disappearance of moles of reactants
Chem. 14.1
Chem 116, Prof. T.L. Heise
7
Chemical Kinetics
Using a graph
such as this
The instantaneous
rate is obtained
using the tangent
of the curve
Chem. 14.1
Chem 116, Prof. T.L. Heise
8
Chemical Kinetics
Sample exercise: Using the graph below,
estimate the instantaneous rate of
disappearance of C4H9Cl at t = 300 s.
Chem. 14.1
Chem 116, Prof. T.L. Heise
9
Chemical Kinetics
10
Sample exercise: Using the graph below,
estimate the instantaneous rate of
disappearance of C4H9Cl at t = 300 s.
1. Draw line at 300 sec
that is tangent to
curve, make a right
triangle
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
Sample exercise: Using the graph below,
estimate the instantaneous rate of
disappearance of C4H9Cl at t = 300 s.
2. DC4H9Cl =
Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
11
Chemical Kinetics
Previous examples were all one to one
relationships, but if the relationship is not
one to one, the coefficients need to be
included.
aA + bB --> cC + dD
Rate = -1 D[A] = -1 D[B] = 1 D[C] = 1 D[D]
a Dt
b Dt c Dt
d Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
12
Chemical Kinetics
13
Sample exercise: The decomposition of N2O5
proceeds according to the equation
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
14
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
1. Rate = -1 N2O5 = 1 DNO2
2 Dt
4 Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
15
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
2. Rate = -4 Dt DN2O5 = 2 Dt DNO2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
16
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
3. Rate = -4 Dt DN2O5 = DNO2
2 Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
17
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
4. Rate = -4 DN2O5 = DNO2
2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
18
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
5. Rate = -2 DN2O5 = DNO2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
19
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
5. Rate = -2 DN2O5 = DNO2
= 2(4.2 x 10-7 M/s)
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
20
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
5. Rate = -2 DN2O5 = DNO2
= 2(4.2 x 10-7 M/s)
= 8.4 x 10-7 M/s
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
21
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
1. Rate = -1 DN2O5 = 1 DO2
2 Dt
1 Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
22
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
2. Rate = -1 Dt DN2O5 = 2 Dt DO2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
23
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
3. Rate = -1 Dt DN2O5 = DO2
2 Dt
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
24
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
4. Rate = -1 DN2O5 = DO2
2
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
25
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
5. Rate = -1/2 DN2O5 = DO2
= 1/2(4.2 x 10-7 M/s)
Chem. 14.1
Chem 116, Prof. T.L. Heise
Chemical Kinetics
26
2N2O5(g) --> 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a
particular instant in a reaction vessel is
4.2 x 10-7 M/s, what is the rate of appearance
of (a) NO2; (b) O2
5. Rate = -1/2 DN2O5 = DO2
= 1/2(4.2 x 10-7 M/s)
= 2.1 x 10-7 M/s
Chem. 14.1
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
Reaction rates diminish as the concentrations
of reactions diminish.
Reaction rates generally increase when
reactant concentrations are increased.
An expression, which shows how the rate
depends on the concentrations of reactants,
is called a rate law.
Chem. 14.2
Chem 116, Prof. T.L. Heise
27
Dependence of Rate on Conc.
An expression, which shows how the rate
depends on the concentrations of reactants,
is called a rate law.
NH4+(aq) + NO2(aq) --> N2(g) + 2H2O(l)
Chem. 14.2
Chem 116, Prof. T.L. Heise
28
Dependence of Rate on Conc.
An expression, which shows how the rate
depends on the concentrations of reactants,
is called a rate law.
NH4+(aq) + NO2(aq) --> N2(g) + 2H2O(l)
Rate = k[NH4+][NO2-]
* the constant k in the rate law is called the
rate constant.
Chem. 14.2
Chem 116, Prof. T.L. Heise
29
Dependence of Rate on Conc.
The rate laws for most reactions have the
general form
rate = k[reactant 1]m[reactant 2]n…
The exponents m and n are called reaction
orders, and their sum is the overall reaction
order.
Notice that the reaction orders do not
necessarily correspond to the coefficients in
the balanced chemical equation.
Chem. 14.2
Chem 116, Prof. T.L. Heise
30
Dependence of Rate on Conc.
31
Sample exercise: What is the reaction order of
the reactant H2 in the following equation:
H2(g) + I2(g) --> 2HI(g)
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
32
Sample exercise: What is the reaction order of
the reactant H2 in the following equation:
H2(g) + I2(g) --> 2HI(g)
1
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
Sample exercise: What are the units of the
rate constant for:
CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g)
Chem. 14.2
Chem 116, Prof. T.L. Heise
33
Dependence of Rate on Conc.
Sample exercise: What are the units of the
rate constant for:
CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g)
units of rate
units of concentration
Chem. 14.2
Chem 116, Prof. T.L. Heise
34
Dependence of Rate on Conc.
Sample exercise: What are the units of the
rate constant for:
CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g)
(M/s)
M3/2
Chem. 14.2
Chem 116, Prof. T.L. Heise
35
Dependence of Rate on Conc.
Sample exercise: What are the units of the
rate constant for:
CHCl3(g) + Cl2(g) --> CCl4(g) + HCl(g)
M-1/2 s-1
Chem. 14.2
Chem 116, Prof. T.L. Heise
36
Dependence of Rate on Conc.
37
Using Initial Rates to Determine Rate Laws:
The rate law for any chemical reaction must
be determined experimentally.
- Rate laws of 0, changing concentrations
will have no effect on rate
- Rate laws of 1, changing concentrations
will produce proportional changes
- Rate laws of 2, changing concentrations
will produce exponential changes
** rate depends on conc. NOT rate constants
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
38
Sample exercise: A particular reaction was found to
depend on the concentration of the hydrogen ion,
[H+]. The initial rates varied as a function of [H+]
as follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7
(a) What is the order of the reaction in [H+]?
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
39
Sample exercise: A particular reaction was found to
depend on the concentration of the hydrogen ion,
[H+]. The initial rates varied as a function of [H+]
as follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7
(a) What is the order of the reaction in [H+]?
-1 , rate is inversely
proportional
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
40
Sample exercise: A particular reaction was found to
depend on the concentration of the hydrogen ion,
[H+]. The initial rates varied as a function of [H+]
as follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7
(b) Predict the initial reaction rate when
[H+] = 0.400M.
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
41
Sample exercise: A particular reaction was found to
depend on the concentration of the hydrogen ion,
[H+]. The initial rates varied as a function of [H+]
as follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7
(b) Predict the initial reaction rate when
[H+] = 0.400M.
-should divide in half again
Chem. 14.2
Chem 116, Prof. T.L. Heise
Dependence of Rate on Conc.
42
Sample exercise: A particular reaction was found to
depend on the concentration of the hydrogen ion,
[H+]. The initial rates varied as a function of [H+]
as follows:
[H+] (M)
0.0500
0.100
0.200
Initial rate (M/s) 6.4 x 10-7 3.2 x 10-7 1.6 x 10-7
(b) Predict the initial reaction rate when
[H+] = 0.400M.
8.0 x 10-8
Chem. 14.2
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
43
Rate laws can be converted into equations that tell
us what the concentrations of the reactants or
products are at any time during the course of a
reaction.
Two types: first order overall
second order overall
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
44
First Order - rate depends on the concentration of
a single reactant raised to the first power
A --> products
rate = - D[A] = k[A]
Dt
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
45
First Order - transform previous equation into an
equation that relates the concentration of A at the
start of the reaction, [A]0, to its concentration at
any other time t, [A]t
ln [A]t - ln [A]0 = -kt
ln [A]t = -kt
[A]0
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
46
First Order - transform previous equation into an
equation that relates the concentration of A at the
start of the reaction, [A]0, to its concentration at
any other time t, [A]t
ln [A]t - ln [A]0 = -kt
ln [A]t = -kt + ln[A]0
y = mx + b
* notice the slope of the line is -k, the rate constant
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
Equation:
47
ln [A]t = -kt + ln[A]0
can be used to determine
1) the concentration of a reactant remaining
at any time after the reaction has begun
2) the time required for a given fraction of a
sample to react
3) the time required for a reactant
concentration to reach a certain level
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
48
Sample exercise: The decomposition of dimethyl
ether, (CH3)2O, at 510°C is a first-order process
with a rate constant of 6.8 x 10-4 s-1.
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s?
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
49
Sample exercise: The decomposition of dimethyl
ether, (CH3)2O, at 510°C is a first-order process
with a rate constant of 6.8 x 10-4 s-1.
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s?
1. ln [A]t = -kt + ln[A]0
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
50
Sample exercise: The decomposition of dimethyl
ether, (CH3)2O, at 510°C is a first-order process
with a rate constant of 6.8 x 10-4 s-1.
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s?
1. ln [A]t = -kt + ln[A]0
ln [A]1420 = -(6.8x 10-4 s-1)(1420) + ln[135]0
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
51
Sample exercise: The decomposition of dimethyl
ether, (CH3)2O, at 510°C is a first-order process
with a rate constant of 6.8 x 10-4 s-1.
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s?
1. ln [A]t = -kt + ln[A]0
ln [A]1420 = -(6.8x 10-4 s-1)(1420) + ln[135]0
ln [A]1420 = -0.9656 + 4.905
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
52
Sample exercise: The decomposition of dimethyl
ether, (CH3)2O, at 510°C is a first-order process
with a rate constant of 6.8 x 10-4 s-1.
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
If the initial pressure of (CH3)2O is 135 torr, what is
its partial pressure after 1420 s?
1. ln [A]1420 = -0.9656 + 4.905
[A]1420 = e3.940
[A]1420 = 51 torr
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
Half Lives are a special case of first order
reactions.
ln 1/2[A]0 = -kt1/2
[A]0
* initial concentration falls out of equation!!
ln 1/2 = -kt1/2
t1/2 = 0.693
k
Chem. 14.3
Chem 116, Prof. T.L. Heise
53
Change of Conc. With Time
Sample exercise: Calculate t1/2 for the following
reaction
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
Chem. 14.3
Chem 116, Prof. T.L. Heise
54
Change of Conc. With Time
Sample exercise: Calculate t1/2 for the following
reaction
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
1. t1/2 = 0.693
k
Chem. 14.3
Chem 116, Prof. T.L. Heise
55
Change of Conc. With Time
Sample exercise: Calculate t1/2 for the following
reaction
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
1. t1/2 = 0.693 = 0.693
k
6.8 x 10-4 s-1
Chem. 14.3
Chem 116, Prof. T.L. Heise
56
Change of Conc. With Time
Sample exercise: Calculate t1/2 for the following
reaction
(CH3)2O(g) --> CH4(g) + H2(g) + CO(g)
1. t1/2 = 0.693 = 0.693
k
6.8 x 10-4 s-1
= 1.02 x 103 s
Chem. 14.3
Chem 116, Prof. T.L. Heise
57
Change of Conc. With Time
58
Second Order Reactions: rate depends on the
reactant concentration raised to the second power
or on the concentrations of two different
reactants, each raised to the first power.
Rate = - D[A] = k[A]2
Dt
1 = kt + 1
[A]t
[A]0
y = mx + b
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
59
Second Order Reactions: a plot of 1/[A]t versus t
will yield a straight line where the slope equals k
** one way to distinguish between first and second
order reactions is to look at the graphs.
A downward straight line after plotting
ln[A] vs t will always be first order, slope
equals -k
An upward straight line after plotting
1/[A] vs t will always be second order,
slope equals k
Chem. 14.3
Chem 116, Prof. T.L. Heise
Change of Conc. With Time
60
Second Order Reactions: half lives of elements or
compounds that decay at a second order rate will
be dependent on the initial concentration:
t1/2 = 1
k[A]0
*practice exercise page 524
Chem. 14.3
Chem 116, Prof. T.L. Heise
Temperature and Rate
61
The rates of most chemical reactions increase as the
temperature rises
- due to an increase in the rate constant with
increasing temperature
- explained by the collision model theory
molecules
must collide to react
the greater number of collisions occuring per
second, the greater the rate
as concentration increases, the number of
collisions increases
Chem. 14.4
Chem 116, Prof. T.L. Heise
Temperature and Rate
62
The rates of most chemical reactions increase as the
temperature rises
- due to an increase in the rate constant with
increasing temperature
- explained by the collision model theory
molecules
move faster, they collide more
forcefully and more frequently, increasing
reaction rates
for a reaction to occur, though, more is required
than simply a collision
Chem. 14.4
Chem 116, Prof. T.L. Heise
Temperature and Rate
63
Collision theory continuedmolecules must possess a certain minimum amount of
energy in order to react
kinetic energy of the molecules can be used to stretch,
bend, and ultimately break bonds, leading to chemical
reactions
if molecules are moving too slowly, with too little
kinetic energy, they merely bounce off one another
without changing
minimum energy required to initiate a chemical
reaction is called the activation energy, Ea
Chem. 14.4
Chem 116, Prof. T.L. Heise
Temperature and Rate
Activation Energy
Chem. 14.4
Chem 116, Prof. T.L. Heise
64
Temperature and Rate
Potential Energy Diagram
Chem. 14.4
Chem 116, Prof. T.L. Heise
65
Temperature and Rate
66
Collision Theory continued…
not every collision in which reactants have an
energy Ea or greater results in reaction
they must be oriented in a certain way for the
collision to lead to a reaction
Arrhenius Equation: reaction rate data depends on
1) fraction of molecules possessing Ea
2) number of collisions per second
3) fraction of collisions with correct orientation
Chem. 14.4
Chem 116, Prof. T.L. Heise
Temperature and Rate
Arrhenius Equation: reaction rate data
depends on
1) fraction of molecules possessing Ea
2) number of collisions per second
3) fraction of collisions with correct orientation
k = Ae-Ea/RT
ln k1 = Ea 1 - 1
k2 R T2 T1
Chem. 14.4
Chem 116, Prof. T.L. Heise
67
Reaction Mechanisms
68
The process by which a reaction occurs is
called the reaction mechanism.
- describes in great detail the order in
which bonds are broken and formed and
the changes in relative positions of the
atoms in the course of the reaction
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
69
Elementary steps - a process that occurs in a
single step
if a single molecule was involved, it is
unimolecular
if two molecules were involved, it is
bimolecular
if three molecules were involved, it is
termolecular
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
70
Multistep Mechanisms - a process that occurs
in a single step
* the elementary steps in a multistep
mechanism must always add to give the
chemical equation of the overall process
- a molecule that is produced in one step
and consumed in the very next step is
considered to be an intermediate
intermediates are not activated complexes
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
71
Sample exercise: For the following reaction of Mo(CO)6
Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO
the proposed mechanism
Mo(CO)6 --> Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3
(a) Is the proposed mechanism consistent with the
equation for the overall reaction?
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
72
Sample exercise: For the following reaction of Mo(CO)6
Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO
the proposed mechanism
Mo(CO)6 --> Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3
(a) Is the proposed mechanism consistent with the
equation for the overall reaction?
Yes, the two equations add up correctly
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
73
Sample exercise: For the following reaction of Mo(CO)6
Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO
the proposed mechanism
Mo(CO)6 --> Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3
(b) Identify the intermediate or intermediates
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
74
Sample exercise: For the following reaction of Mo(CO)6
Mo(CO)6 + P(CH3)3 --> Mo(CO)5P(CH3)3 + CO
the proposed mechanism
Mo(CO)6 --> Mo(CO)5 + CO
Mo(CO)5 + P(CH3)3 --> Mo(CO)5P(CH3)3
(b) Identify the intermediate or intermediates
Mo(CO)5
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
75
The rate law for a reaction can be determined from its
mechanisms, so our next step is to arrive at reaction
mechanisms that lead to rate laws that are consistent
with those observed experimentally.
If a reaction is an elementary step, we know the rate
law automatically, and it is first order
A --> products
rate = k[A]
If a reaction is a bimolecular elementary step, the
rate law is second order
A + B --> products
rate = k[A][B]
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
,
Chem. 14.5
Chem 116, Prof. T.L. Heise
76
Reaction Mechanisms
77
Sample exercise:
Consider the following reaction:
2NO(g) + Br2(g) --> 2NOBr(g)
(a) Write the rate law for the reaction, assuming it involves
a single elementary step.
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
78
Sample exercise:
Consider the following reaction:
2NO(g) + Br2(g) --> 2NOBr(g)
(a) Write the rate law for the reaction, assuming it involves
a single elementary step.
Rate =k[NO]2[Br2]
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
Sample exercise:
Consider the following reaction:
2NO(g) + Br2(g) --> 2NOBr(g)
(b) Is a single-step mechanism likely for this reaction?
Chem. 14.5
Chem 116, Prof. T.L. Heise
79
Reaction Mechanisms
80
Sample exercise:
Consider the following reaction:
2NO(g) + Br2(g) --> 2NOBr(g)
(b) Is a single-step mechanism likely for this reaction?
No, because termolecular reactions are very
rare
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
81
Often one of the steps is much slower than the others. The
overall rate of a reaction cannot exceed the rate of the
slowest elementary step of its mechanisms.
Slowest Step is the rate-determining step!
- the rate determining step governs the rate law for the
overall reaction
-mechanisms with an initial fast step, can be solved for
the concentration of an intermediate be assuming that an
equilibrium is established in the fast step
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
82
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
83
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
Rate = k1 [Br2]
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
84
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
Rate = k1 [Br2]
Rate = k-1[Br]2
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
85
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
Rate = k1 [Br2]
Rate = k-1[Br]2
*rates are equal so ...
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
86
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
k1 [Br2] = k-1[Br]2
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
87
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
k1 [Br2] = [Br]2
k-1
Chem. 14.5
Chem 116, Prof. T.L. Heise
Reaction Mechanisms
88
Practice exercise: The first step of a mechanism involving
the reaction of bromine is
Br2(g) <----> 2Br(g) (fast equilibrium)
What is the expression relating the concentration of
Br(g) to that of Br2(g)?
k1 [Br2] = [Br]
k-1
Chem. 14.5
Chem 116, Prof. T.L. Heise
Catalysis
89
A catalyst is a substance that changes the
speed of a chemical reaction without undergoing a
permanent chemical change itself in the process.
Catalysts are very common
•reactions in the body
•in the atmosphere
•in the oceans
•in industrial chemistry
Chem. 14.6
Chem 116, Prof. T.L. Heise
Catalysis
90
A catalyst that is present in the same phase as the
reacting molecules is a homogeneous catalyst
- the rate constant k is determined by the
activation energy, Ea, and the frequency
factor, A.
- a catalyst may affect the rate of a reaction by
altering the value for either Ea or A.
- the most dramatic catalytic effects come
from lowering Ea
- lowers Ea by providing a completely
different mechanism for the reaction
Chem. 14.6
Chem 116, Prof. T.L. Heise
Catalysis
91
Homogeneous catalyst
Chem. 14.6
Chem 116, Prof. T.L. Heise
Catalysis
92
A catalyst that is different in phase as the reacting
molecules is a heterogeneous catalyst
- usually as a solid in contact with either
gaseous reactants or with reactants in the
liquid solution
- often composed of metals or metal oxides
- catalyzed reactions occurs on the surface,
special methods are often used to prepare
catalysts so that they have very large surface
areas
- the initial step is ADSORPTION
Chem. 14.6
Chem 116, Prof. T.L. Heise
Catalysis
93
Enzymes
- many of the most interesting and important
examples of catalysis involve reactions within
living systems.
- A large number of marvelously efficient
biological catalysts known as enzymes are
necessary for many of these reaction to occur
at suitable rates
- most enzymes are large protein molecules
with molecular weights ranging from 10,000
to about 1 million amu
Chem. 14.6
Chem 116, Prof. T.L. Heise