Chapter 14 Chemical Kinetics
seconds
minutes
months
4Fe(s) + 3O2 (g) ® 2Fe2O3 (s)
Dr. Nick Blake
Ventura Community College
http://chem.aellumis.org
Kinetics
• Kinetics is the study of the factors that affect the
speed of a reaction and the mechanism by which a
reaction proceeds
• Experimentally it is shown that there are 4 factors
that influence the speed of a reaction:
 nature of the reactants
 temperature
 catalysts
 concentration
2
Defining Rate
• Rate is how much a quantity changes in a given period of
•
time
The speed you drive your car is a rate – the distance your
car travels (miles) in a given period of time (1 hour)
x final - xinitial distance travelled
Speed =
=
t final - tinitial
time interval
Dx
=
Dt
3
Defining Reaction Rate
• the rate of a chemical reaction is measured as the change
•
of concentration of a reactant (or product) in a given
period of time interval
To make the rate > 0, for reactants, a negative sign is
placed in front of the definition
D concentration
Rate =
D time
D [product]
D [reactant]
Rate =
=D time
D time
4
Reaction Rate: Changes Over
Time
• as time goes on, the rate of a reaction generally slows down
 because the concentration of the reactants decreases
• at some time the reaction stops, either because the reactants
run out or because the system has reached equilibrium
5
at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
(
)
[]A2 ]-2 [-C[ ]A1 ]1 )
(
DD[[CA]] [ C
Rate
Rate==== Dt
Dt
( t 2( t-2 t-1 )t1 )
(44--08)) = 0.25
(
Rate
=
Rate = (16 - 0 ) = 0.25
(16 - 0 )
(
)
Z []1X ]1 )
DD[[ZX] ] [ Z([]2X-]2[ Rate
Rate==== DtDt
( t 2 (-t 2t1-) t1 )
7 -08) )
((10.0625
==0.0625
(16--00))
(16
Rate
Rate==-
6
at t = 16
[A] = 4
[B] = 4
[C] = 4
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2
( (
) )
( (
) )
- ][ C
- A
D [DC[ ]A ] [ C] [f A
f ]i [ ]i
Rate = ==DtDt
t f - ttfi - t i
[ Z ][f X-][fZ-]i[ X]i
DD[ Z[ X
]
]
Rate
Rate == == DtDt
t f -ttfi - t i
2 -4 )4 )
6( (
0.125
Rate = = =0.125
- 16
( 32- 16
( 32
))
2( 6--1)7 )
(
Rate
0.0625
Rate == ==0.0625
( 32--1616) )
( 32
( ( ) )
( ( ) )
7
at t = 32
[A] = 2
[B] = 2
[C] = 6
at t = 32
[X] = 6
[Y] = 6
[Z] = 2
at t = 48
[A] = 0
[B] = 0
[C] = 8
at t = 48
[X] = 5
[Y] = 5
[Z] = 3
((
))
- -C A
A]] [ C[]A
D[C
f ] f [ []i ]i
Rate
Rate= === Dt
Dt
t f t-i t i
tf Rate
Rate= =-
( 80 - 62)
( 48 - 32 )
(( ))
= 0.125
( (
) )
[ Z ][f X-][fZ-]i[ X]i
DD[ Z[ X
]
]
Rate
Rate == == DtDt
t f -ttfi - t i
( ( ) )
3( 5- -2 )6 )
(
Rate
0.0625
Rate == ==0.0625
( 48--3232) )
( 48
8
Reaction Rate and Stoichiometry
• in most reactions, the coefficients of the balanced equation are not all the
same
H2 (g) + I2 (g)  2 HI(g)
• for these reactions, the change in the number of molecules of one
substance is a multiple of the change in the number of molecules of
another
 for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be
used and 2 moles of HI made
 therefore the rate of change will be different
• in order to be consistent, the change in the concentration of each
substance is multiplied by 1/coefficient
D[H 2 ]
D[I 2 ]
æ 1 ö D[HI]
Rate = == +ç ÷
Dt
Dt
è 2 ø Dt
9
Average Rate
• the average rate is the change in measured
concentrations in any particular time period
 linear approximation of a curve
• the larger the time interval, the more the average rate
deviates from the instantaneous rate
10
H2
I2
HI
t (s)
[H2]t, M
[HI]t, M
0.000
1.000
0.000
10.000
0.819
0.362
20.000
0.670
30.000
0.549
HI
Avg. Rate, M/s
Avg. Rate, M/s
-Δ[H2]/Δt
½ Δ[HI]/Δt
[ HI ]t = -2 × ([ H 2 ]t - [ H 2 ]0 )
0.0181
0.0181 ö
D[H
-0.181 M
æ 0.819 - 1.000
2]
rate = = -ç
=
÷ø
è
Dt
10.000
0.000
10.000 s
0.660
0.0149
0.0149
0.902
= 0.0181 Ms
0.0121
0.0121
average rate between t=0s and t=10s
40.000
0.449
1.102
0.0100
0.0100
50.000
0.368
1.264
0.0081
0.0081
60.000
0.301
1.398
0.0067
0.0067
70.000
0.247
1.506
0.0054
0.0054
80.000
0.202
1.596
0.0045
0.0045
90.000
0.165
1.670
0.0037
0.0037
11
100.000
0.135
1.730
0.0030
0.0030
Concentration vs. Time for H2 + I2 --> 2HI
2.000
1.800
1.600
average rate = - slope of the
line connecting the [H2] points;
= ½ slope of the line for [HI]
concentration, (M)
1.400
1.200
the average rate for the
first 80
10 s is 0.0108
40
0.0181 M/s
0.0150
1.000
[H2], M
[HI], M
0.800
0.600
0.400
0.200
0.000
0.000
20.000
Tro, Chemistry: A Molecular Approach
40.000
t (s)
60.000
80.000
100.000
12
Instantaneous Rate
• the instantaneous rate at a particular t is the change in
concentration at that t
 Slope of the concentration vs time graph at t
• determined by taking the slope of a line tangent to the
curve at that particular point
 first derivative of the function
13
H2 (g) + I2 (g)  2 HI (g)
Using [H2], the
instantaneous rate at 50
s is:
D[H 2 ]
Dt
-0.28 M
=40 s
= 0.0070 Ms
rate = -
Using [HI], the
instantaneous rate at 50
s is:
1 D[HI ]
2 Dt
æ 1 ö 0.56 M
=ç ÷
è 2 ø 40 s
rate =
= 0.0070
M
s 14
Example Problem
For the reaction given, the [I-] changes from 1.000 M to 0.868 M in the
first 10 s. Calculate the average rate in the first 10 s and the Δ[H+].
H2O2 (aq) + 3 I-(aq) + 2 H+(aq)  I3- (aq) + 2 H2O(l)
Solve the equation for
the Rate (in terms of the
change in concentration
of the given reactant
here (I-))
æ 1 ö D[I ]
æ 1 ö ( 0.868 M -1.000 M )
Rate = - ç ÷
= -ç ÷
è 3 ø Dt
è 3ø
10 s
Rate = 4.40 ´10 -3 Ms
Rewrite the Rate in
1 ö D[H + ]
æ
terms of the change in
Rate = - ç ÷
= 4.40 ´ 10 -3 Ms
è 2 ø Dt
the concentration for the
product to find) (H+) and D[H + ] = -2 ( Rate ) × Dt
solve for the unknown
(
)
D[H + ] = -2 4.40 ´ 10 -3 Ms ×10s = - 8.80 ´ 10 -2 M
Measuring Reaction Rate
• In order to measure the reaction rate you need to be able to measure
the concentration of at least one component in the mixture at many
points in time
• There are two ways of approaching this problem
 (1) for reactions that are complete in less than 1 hour, it is best to use continuous
monitoring of the concentration
 (2) for reactions that happen over a very long time, sampling of the mixture at
various times can be used
 when sampling is used, often the reaction in the sample is stopped
by a quenching technique
16
Continuous Monitoring
• polarimetry – measuring the change in the degree of rotation
of plane-polarized light caused by one of the components
over time
• spectrophotometry – measuring the amount of light of a
particular wavelength absorbed by one component over time
• total pressure – the total pressure of a gas mixture is
stoichiometrically related to partial pressures of the gases in
the reaction
17
Sampling
• gas chromatography can measure the concentrations of
various components in a mixture
 for samples that have volatile components
 separates mixture by adherence to a surface
• drawing off periodic aliquots from the mixture and doing
quantitative analysis
 titration for one of the components
 gravimetric analysis
18
Factors Affecting Reaction Rate
Nature of the Reactants
• nature of the reactants means what kind of reactant molecules and what
physical condition they are in.
 small molecules tend to react faster than large molecules;
 gases tend to react faster than liquids which react faster than solids;
 powdered solids are more reactive than “blocks”
 more surface area for contact with other reactants
 certain types of chemicals are more reactive than others
 e.g., the activity series of metals
 ions react faster than molecules
 no bonds need to be broken
19
Factors Affecting Reaction Rate
Temperature
• Increasing temperature increases reaction rate
 chemist’s rule of thumb - for each 10°C rise in temperature, the
speed of the reaction doubles
• there is a mathematical relationship between the
absolute temperature and the speed of a reaction
discovered by Svante Arrhenius which will be examined
later
20
Factors Affecting Reaction Rate
Catalysts
Catalysts are substances which affect the speed of a reaction
without being consumed
• most catalysts are used to speed up a reaction, these are
called positive catalysts
 catalysts used to slow a reaction are called negative catalysts
• homogeneous = present in same phase
• heterogeneous = present in different phase
• how catalysts work will be examined later
21
Factors Affecting Reaction Rate
Reactant Concentration
• generally, the larger the concentration of reactant
molecules, the faster the reaction
• increases the likelihood of reactant molecules bumping into
•
each other
concentration of gases depends on the partial pressure of the
gas
 higher pressure = higher concentration
• concentration of solutions depends on the solute to
solution ratio (molarity)
22
The Rate Law
• the Rate Law of a reaction is the mathematical relationship
between the rate of the reaction and the concentrations of
the reactants
 and homogeneous catalysts as well
• the rate of a reaction is directly proportional to the
concentration of each reactant raised to a power
• for the reaction aA + bB  products the rate law would
have the form given below
 n and m are called the orders for each reactant
 k is called the rate constant
n
m
Rate = k[A] [B]
23
Reaction Order
• the exponent on each reactant in the rate law is called the order with
respect to that reactant
• the sum of the exponents on the reactants is called the order of the
reaction
• The rate law for the reaction:
2 NO(g) + O2(g)  2 NO2(g)
Rate = k[NO]x[O2]y
When it is measured experimentally it is found to be
Rate = k[NO]2[O2]1=k[NO]2[O2]
The reaction is:
second order with respect to [NO] (x=2),
first order with respect to [O2] (y=1),
and third order overall = (x + y = 2 + 1 = 3)
24
Sample Rate Laws
Reaction
Rate Law
CH3CN ® CH 3NC
Rate = k[CH 3CN]
CH3CHO ® CH 4 + CO
Rate = k[CH 3CHO] 3/2
2 N 2O5 ® 4 NO 2 + O 2
Rate = k[N 2O5]
H2 + I 2 ® 2 HI
Rate = k[H 2][I 2]
+3
Tl +
Hg 2+2
® Tl + 2 Hg
+1
+2
Rate = k[Tl +3][Hg 2+2][Hg +2]-1
The reaction is autocatalytic, because a product affects the rate.
Hg2+ is a negative catalyst, increasing its concentration slows the reaction.
25
Reactant Concentration vs. Time
A  Products
26
Half-Life
• the half-life, t1/2, of a
reaction is the length of time
it takes for the
concentration of the
reactants to fall to ½ its
initial value
• the half-life of the reaction
depends on the order of the
reaction
Tro, Chemistry: A Molecular Approach
27
Zero Order Reactions
• Rate = k[A]0 = k
•
•
•
•
•
 constant rate reactions
[A] = -kt + [A]0
graph of [A] vs. time is straight line with
slope = -k and y-intercept = [A]0
t ½ = [A0]/2k
[A]0
when Rate = M/sec, k = M/sec
[A]
time
28
First Order Reactions
• Rate = k[A]
• ln[A] = -kt + ln[A]0
rate = -
d[A]
= k[A]
dt
or
[ A]
t
d[A]
= -k ò dt
ò
[A]
[ A]o
0
• graph ln[A] vs. time gives straight line with slope = -k and
y-intercept = ln[A]0
 used to determine the rate constant
• t½ = 0.693/k
• the half-life of a first order reaction is constant
• the when Rate = M/sec, k = sec-1
29
ln[A]0
ln[A]
time
30
Half-Life of a First-Order Reaction
Is Constant
31
Rate Data for
C4H9Cl + H2O  C4H9OH + HCl
Time (sec)
0.0
[C4H9Cl], M
0.1000
50.0
100.0
150.0
0.0905
0.0820
0.0741
200.0
300.0
400.0
500.0
0.0671
0.0549
0.0448
0.0368
800.0
10000.0
0.0200
0.0000
32
C4H9Cl + H2O  C4H9OH + 2 HCl
Concentration vs. Time for the Hydrolysis of C4H9Cl
0.12
0.1
concentration, (M)
0.08
0.06
0.04
0.02
0
0
200
400
600
time, (s)
800
1000
33
C4H9Cl + H2O  C4H9OH + 2 HCl
Rate vs. Time for Hydrolysis of C
4 H9 Cl
2.5E-04
Rate, (M/s)
2.0E-04
1.5E-04
1.0E-04
5.0E-05
0.0E+00
0
100
200
300
400
500
600
700
800
time, (s)
34
C4H9Cl + H2O  C4H9OH + 2 HCl
LN([C4H9Cl]) vs. Time for Hydrolysis of C4H9Cl
0
-0.5
slope =
-2.01 x 10-3
k=
2.01 x 10-3 s-1
LN(concentration)
-1
-1.5
-2
-2.5
-3
y = -2.01E-03x - 2.30E+00
-3.5
t1
=
-4
-4.5
0
100
200
300
400
500
600
700
800
2
0.693
=
k
0.693
2.01´10 -3 s -1
= 345 s
time, (s)
35
Second Order Reactions
• Rate = k[A]2
• 1/[A] = kt + 1/[A]0
• graph 1/[A] vs. time gives straight line with slope = k
and y-intercept = 1/[A]0
 used to determine the rate constant
• t½ = 1/(k[A0])
• when Rate = M/sec, k = M-1∙sec-1
Tro, Chemistry: A Molecular Approach
36
1/[A]
l/[A]0
time
Tro, Chemistry: A Molecular Approach
37
Rate Data For
2 NO2  2 NO + O2
Time (hrs.)
0
30
60
Partial
Pressure
NO2, mmHg
100.0
62.5
45.5
ln(PNO2)
4.605
4.135
3.817
1/(PNO2)
0.01000
0.01600
0.02200
90
120
150
35.7
29.4
25.0
3.576
3.381
3.219
0.02800
0.03400
0.04000
180
210
240
21.7
19.2
17.2
3.079
2.957
2.847
0.04600
0.05200
0.05800
38
Rate Data Graphs For
2 NO2  2 NO + O2
Partial Pressure NO2, mmHg vs. Time
100.0
90.0
Pressure, (mmHg)
80.0
70.0
60.0
50.0
40.0
30.0
20.0
10.0
0.0
0
50
100
150
Time, (hr)
200
250
39
Rate Data Graphs For
2 NO2  2 NO + O2
ln(PNO2) vs. Time
4.8
4.6
4.4
4.2
ln(pressure)
4
3.8
3.6
3.4
3.2
3
2.8
2.6
2.4
0
50
100
150
Time (hr)
200
250
40
Rate Data Graphs For
2 NO2  2 NO + O2
1/(PNO2) vs Time
0.07000
Inverse Pressure, (mmHg-1)
0.06000
1/PNO2 = 0.0002( me) + 0.01
0.05000
0.04000
0.03000
0.02000
0.01000
0.00000
0
50
100
150
Time, (hr)
200
250
41
Determining the Rate Law
• can only be determined experimentally
• graphically
 rate = slope of curve [A] vs. time
 if graph [A] vs time is straight line, then exponent on A in
rate law is 0, rate constant = -slope
 if graph ln[A] vs time is straight line, then exponent on A in
rate law is 1, rate constant = -slope
 if graph 1/[A] vs time is straight line, exponent on A in rate
law is 2, rate constant = slope
• initial rates
 by comparing effect on the rate of changing the initial
concentration of reactants one at a time
42
Tro, Chemistry: A Molecular Approach
43
Practice - Complete the Table and Determine the Rate
Equation for the Reaction A  2 Prod
[A], (M)
0.100
[Prod], (M) Time (sec)
0
ln([A])
1/[A]
0
0.067
50
0.050
100
0.040
150
0.033
200
0.029
250
44
Practice - Complete the Table and Determine the Rate
Equation for the Reaction A  2 Prod
[A], (M) [Prod], (M) Time (sec)
ln([A])
1/[A]
0.100
0
0
-2.3
10
0.067
0.066
50
-2.7
15
0.050
0.100
100
-3.0
20
0.040
0.120
150
-3.2
25
0.033
0.134
200
-3.4
30
0.029
0.142
250
-3.5
35
Tro, Chemistry: A Molecular Approach
45
[A] vs. Time
0.12
concentration, M
0.1
0.08
0.06
0.04
0.02
0
0
50
100
150
200
250
time, (s)
Tro, Chemistry: A Molecular Approach
46
LN([A]) vs. Time
-2
-2.2
Ln(concentration)
-2.4
-2.6
-2.8
-3
-3.2
-3.4
-3.6
-3.8
0
50
100
150
200
250
time, (s)
Tro, Chemistry: A Molecular Approach
47
1/([A]) vs. Time
y = 0.1x + 10
40
inverse concentration, M
-1
35
30
25
20
15
10
5
0
0
50
100
150
200
250
time, (s)
Tro, Chemistry: A Molecular Approach
48
Practice - Complete the Table and
Determine the Rate Equation for the
Reaction A  2 Prod
the reaction is second order,
Tro, Chemistry: A Molecular Approach
-D[A]
Rate =
= 0.1 [A]2
Dt
49
The reaction SO2Cl2(g)  SO2(g) + Cl2(g) is first order with a rate
constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the
[SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M
Given:
[SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1
Find:
Concept Plan:
[SO2Cl2]
Relationships:
for a 1st order process : ln[A] = kt + ln[A]0
ln[SO2Cl2 ] = kt + ln[SO 2Cl2 ]0
Solution:
[SO2Cl2]0, t, k
[SO2Cl2]
ln[SO 2Cl2 ] = ( 2.90 ´ 10 -4 s-1 ) ( 865 s ) + ln ( 0.0225 )
ln[SO 2Cl2 ] = -0.251 - 3.79 = -4.04
[SO2Cl2 ] = e(-4.04) = 0.0175 M
Check: the new concentration is less than the original, as
expected
Initial Rate Method
• another method for determining the order of a reactant
is to see the effect on the initial rate of the reaction
when the initial concentration of that reactant is
changed
• for multiple reactants, keep initial concentration
of all reactants constant except one
• zero order = changing the concentration has no
effect on the rate
• first order = the rate changes by the same factor as
the concentration
• doubling the initial concentration will double the rate
• second order = the rate changes by the square of
the factor the concentration changes
• doubling the initial concentration will quadruple the rate
51
Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Write the general
rate law
Take the ratios of 2
experiments where
one reactant is
changing but other
reactants are fixed
Exp.
Initial
Initial Rate
Initial
Expt.
Number
(M)
(M/s)
[NO22],
], (M)
(M) [CO],
Number [NO
[CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
Rate = k[NO2 ]n [CO]m
0.40
0.10
0.033
Comparing Exp #1 and Exp #2, the [NO2]
changes but the [CO] does not
n
rate2 0.0082
k [NO]n2 [CO]2m [0.2]n [0.1]m é 0.2 ù
n
=
»4= ×
=
×
=
=
2
rate1 0.0021
k [NO]1n [CO]1m [0.1]n [0.1]m êë 0.1 úû
52
Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Repeat for the
other reactants
Expt.
Exp.
Initial
Initial Rate
Initial
Number [NO2], (M) [CO], (M)
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
m
[CO]3 0.20
=
[CO]2 0.10
æM[CO]3 ö
Rate
3
Rate
0.0083
=
3
çè [CO]
=
= 22 ÷ø
Rate 2
M
2 =1
m
m=0
Rate2
0.0082
M
M
s
»1
s
53
Determine the rate law and rate constant for the
reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Substitute the
exponents into the
general rate law to
get the rate law for
the reaction
Expt.
Number
Initial [NO2],
(M)
Initial
[CO], (M)
Initial Rate
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
n0
m
2
n = 2, mRate
= 0 =Rate
= 2k][NO
k[NO
[CO]
2 ] [CO]
2
Rate = k[NO 2 ]
54
Ex 13.2 – Determine the rate law and rate constant
for the reaction NO2(g) + CO(g)  NO(g) + CO2(g)
given the data below.
Substitute the
concentrations
and rate for any
experiment into
the rate law and
solve for k
Exp.
Number
Initial [NO2],
(M)
Initial
[CO], (M)
Initial Rate
(M/s)
1.
0.10
0.10
0.0021
2.
0.20
0.10
0.0082
3.
0.20
0.20
0.0083
4.
0.40
0.10
0.033
Rate = k[NO 2 ]2
for Exp 1
0.0021
M
s
= k ( 0.10 M )
2
0.0021 M s
-1
-1
k=
=
0.21
M
·
s
0.01 M 2
55
Practice - Determine the rate law and rate constant for
the reaction NH4+1 + NO2-1  N2 + 2 H2O
given the data below.
Expt.
Initial
Initial
+
No. [NH4 ], M [NO2 ], M
Initial
Rate,
(x 10-7),
M/s
1
0.0200
0.200
10.8
2
0.0600
0.200
32.3
3
0.200
0.0202
10.8
4
0.200
0.0404
21.6
56
Practice - Determine the rate law and rate constant for
the reaction NH4+1 + NO2-1  N2 + 2 H2O
given the data below.
Expt.
No.
Initial
[NH4+], M
Initial
[NO2-], M
Initial Rate,
(x 10-7), M/s
1
0.0200
0.200
10.8
2
0.0600
0.200
32.3
3
0.200
0.0202
10.8
4
0.200
0.0404
21.6
Expt 2 0.0600
For [NH 4 ],
=
=3
Expt 1 0.0200
Expt 2 32.3 ´10 -7
Rate,
=
=3
7
Expt 1 10.8 ´10
Rate Factor = [NH +4 ]n 3 = 3n
\ n = 1, first order
+
Rate = k[NH4+]n[NO2-]m
Rate = k[NH 4 + ][NO 2 - ]
for expt 1
10.8 ´ 10-7 M s = k (0.0200 M )(0.200 M )
10.8 ´10-7 M s
k=
= 2.70 ´ 10 - 4 M -1 · s -1
4.00 ´ 10-3 M 2
Expt 4 0.0404
For [NO 2 ],
=
=2
Expt 3 0.0202
Expt 4 21.6 ´10 - 7
Rate,
=
=2
7
Expt 3 10.8 ´10
Rate Factor = [NO -2 ]m 2 = 2 m
\ m = 1, first order
57
-
Collision Theory
• Explains how reactions occur and why reaction rates differ for different
reactions
• Reactions occur when reactant atoms/molecules/ions collide (elementary
steps)
• Some reactions occur in several elementary steps (reaction mechanism)
• Some collisions are effective when the geometry of the reactants is
favorable, and some are not
H2 + I2  2HI
58
Effective Collisions
Orientation Effect
59
Collision Theory
• Only reactive collisions can bring about chemical change
• Reactive collisions must have exceed the activation energy Ea which
goes toward breaking pre-existing bonds
æ -RTEa ö
k = Açç e ÷÷
è
ø
60
Collision Theory
• The rate for an elementary step aA +bBproducts
E
- a
1 D[A]
a
b
a
b
RT
= Ae [ A ] [ B ] = k [ A ] [ B ]
a Dt
Probability of aA and bB
occupying the same point
in space
Aka Exponential factor
# favorable collisions / s
Aka Frequency factor
Fraction of collisions with
enough energy to react
61
The Effect of Temperature on Rate
• changing the temperature changes the rate constant of the
•
rate law
Arrhenius investigated this relationship and showed that:
æ -RTEa
k = Açç e
è
ö
÷
÷
ø
where T is the temperature in Kelvin
R is the gas constant in energy units, 8.314 J/(mol∙K)
A is a factor called the frequency factor (has a weak T dependence)
Ea is the activation energy, the extra energy needed to start the
molecules reacting
62
63
Activation Energy and the
Activated Complex
• energy barrier to the reaction
• amount of energy needed to convert reactants into the
•
activated complex
• aka transition state
the activated complex is a chemical species with partially
broken and partially formed bonds
• always very high in energy because partial bonds
64
Isomerization of Methyl Isonitrile
methyl isonitrile rearranges to acetonitrile
in order for the reaction to occur,
the H3C-N bond must break; and
a new H3C-C bond form
65
Energy Profile for the Isomerization of
Methyl Isonitrile
66
The Arrhenius Equation:
The Exponential Factor
æ -RTEa
k = Açç e
è
• The exponential factor in the Arrhenius equation is a number
between 0 and 1
• it represents the fraction of reactant molecules with sufficient
energy so they can make it over the energy barrier
•
the higher the energy barrier (larger activation energy), the fewer
molecules that have sufficient energy to overcome it
• That extra energy comes from converting the kinetic energy of
motion to potential energy in the molecule when the molecules
collide
•
•
•
increasing the temperature increases the average kinetic energy
of the molecules
therefore, increasing the temperature will increase the number of
molecules with sufficient energy to overcome the energy barrier
therefore increasing the temperature will increase the reaction
rate
67
ö
÷
÷
ø
Tro, Chemistry: A Molecular Approach
68
Arrhenius Plots
• the Arrhenius Equation can be algebraically solved to give the
following form:
-Ea æ 1 ö
ln(k) =
çè ÷ø + ln ( A )
R T
æ -RTEa ö
k = Aç e ÷
è
ø
y
æ -Ea ö
=ç
× x + ln ( A )
è R ÷ø
y
= m×x + b
• (-8.314 J/mol∙K) x (slope of the line) = Ea, (in J/mol)
• ey-intercept = A, (unit is the same as k)
69
Determine the activation energy and frequency factor for the
reaction O3(g)  O2(g) + O(g) given the following data:
Temp, K
600
700
800
k, M-1∙s-1
3.37 x 103
4.83 x 104
3.58 x 105
Temp, K
1300
1400
1500
k, M-1∙s-1
7.83 x 107
1.45 x 108
2.46 x 108
900
1000
1100
1.70 x 106
5.90 x 106
1.63 x 107
1600
1700
1800
3.93 x 108
5.93 x 108
8.55 x 108
1200
3.81 x 107
1900
1.19 x 109
70
Ex. 13.7 Determine the activation energy and frequency factor for
the reaction O3(g)  O2(g) + O(g) given the following data:
use a spreadsheet to
graph ln(k) vs. (1/T)
71
Determine the activation energy and frequency factor for the
reaction O3(g)  O2(g) + O(g) given the following data:
Ea = m∙(-R)
solve for Ea
A = ey-intercept
solve for A
(
)
J
Ea = (1.12 ´10 4 K ) 8.314 molJ·K = 9.31 ´10 4 mol
kJ
Ea = 93.1 mol
A = e26.8 = 4.36 ´1011
A = 4.36 ´1011 M -1s-1
72
Arrhenius Equation:
Two-Point Form
• if you only have two (T,k) data points, the following forms of the
Arrhenius Equation can be used:
æ k 2 ö Ea æ 1 1 ö
çç - ÷÷
lnçç ÷÷ =
è k1 ø R è T1 T2 ø
æ k1 ö
k1
k1
( R × T1 × T2 ) × ln ç k ÷
R × ln
R × ln
è 2ø
k2
k2
Ea =
=
=
1 1
T1 - T2
(T1 - T2 )
T2 T1
T1 × T2
73
The reaction NO2(g) + CO(g)  CO2(g) + NO(g) has a rate constant
of 2.57 M-1∙s-1 at 701 K and 567 M-1∙s-1 at 895 K. Find the
activation energy in kJ/mol
Given:
Find:
Concept Plan:
T1 = 701 K, k1 = 2.57 M-1∙s-1, T2 = 895 K, k2 = 567 M-1∙s-1
Ea, kJ/mol
T1, k1, T2, k2
Ea
æk ö E æ 1
1 ö
lnçç 2 ÷÷ = a çç - ÷÷
è k1 ø R è T1 T2 ø
Relationships:
-1
-1
Ea
æ 567 M · s ö
lnç
=
Solution: è 2.57 M -1 · s -1 ÷ø 8.314 J
mol· K
5.3965 =
1.45 ´105
(
1 ö
æ 1
ç
÷
701
K
895
K
è
ø
Ea
-4
-1
3
.
0
9
2
´
10
K
J
-1
8.314 mol
·K
J
mol
)
kJ
= 145 mol
= Ea
Check: most activation energies are tens to hundreds of
kJ/mol – so the answer is reasonable
Orientation Factor
• the proper orientation results when the atoms are
•
aligned in such a way that the old bonds can
break and the new bonds can form
the more complex the reactant molecules, the
less frequently they will collide with the proper
orientation
 reactions between atoms generally have p = 1
 reactions where symmetry results in multiple
orientations leading to reaction have p slightly less than
1
• for most reactions, the orientation factor is less
than 1
 for many, p << 1
 there are some reactions that have p > 1 in which an
Tro, Chemistry: A Molecular Approach
electron is transferred without direct collision
75
Reaction Mechanisms
• we generally describe chemical reactions with an
equation listing all the reactant molecules and
product molecules
• but the probability of more than 3 molecules
colliding at the same instant with the proper
orientation and sufficient energy to overcome the
energy barrier is negligible
• most reactions occur in a series of small reactions
involving 1, 2, or at most 3 molecules
• describing the series of steps that occur to
produce the overall observed reaction is called a
reaction mechanism
• knowing the rate law of the reaction helps us
understand the sequence of steps in the
mechanism
Tro, Chemistry:
A Molecular Approach
76
Example of a Reaction Mechanism
•
•
•
Overall reaction:
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
Mechanism:
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
the steps in this mechanism are elementary steps, meaning that
they cannot be broken down into simpler steps and that the
molecules actually interact directly in this manner without any other
steps
77
Elements of a Mechanism
Intermediates
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
•
notice that the HI is a product in Step 1, but then a reactant in Step 2
•
since HI is made but then consumed, HI does not show up in the
overall reaction
•
materials that are products in an early step, but then a reactant in a
later step are called intermediates
78
Molecularity
• the number of reactant particles in an elementary step is
called its molecularity
• a unimolecular step involves 1 reactant particle
• a bimolecular step involves 2 reactant particles
 though they may be the same kind of particle
• a termolecular step involves 3 reactant particles
 though these are exceedingly rare in elementary steps
79
Rate Laws for Elementary Steps
• each step in the mechanism is like its own little reaction – with
its own activation energy and own rate law
• the rate law for an overall reaction must be determined
experimentally
• but the rate law of an elementary step can be deduced from
the equation of the step
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
1) H2(g) + ICl(g)  HCl(g) + HI(g)
2) HI(g) + ICl(g)  HCl(g) + I2(g)
overall
Rate = k1[H2][ICl]
Rate = k2[HI][ICl]
80
Rate Laws of Elementary Steps
Tro, Chemistry: A Molecular Approach
81
Rate Determining Step
• in most mechanisms, one step occurs slower than the other steps
• the result is that product production cannot occur any faster than the
slowest step – the step determines the rate of the overall reaction
• we call the slowest step in the mechanism the rate determining step
 the slowest step has the largest activation energy
• the rate law of the rate determining step determines the rate law of the
overall reaction
82
Another Reaction Mechanism
NO2(g) + CO(g)  NO(g) + CO2(g)
1) NO2(g) + NO2(g)  NO3(g) + NO(g)
2) NO3(g) + CO(g)  NO2(g) + CO2(g)
Rateobs = k[NO2]2
Rate = k1[NO2]2 slow
Rate = k2[NO3][CO] fast
The first step is slower than the
second step because its activation
energy is larger
The first step in this mechanism is
the rate determining step
The rate law of the first step is the
same as the rate law of the overall
reaction
83
Validating a Mechanism
In order to validate (not prove) a mechanism, two conditions
must be met:
• the elementary steps must sum to the overall reaction
• the rate law predicted by the mechanism must be consistent with the
experimentally observed rate law
84
Mechanisms with a Fast Initial Step
• when a mechanism contains a fast initial step, the rate
limiting step may contain intermediates
• when a previous step is rapid and reaches equilibrium, the
forward and reverse reaction rates are equal – so the
concentrations of reactants and products of the step are
related
 and the product is an intermediate
• substituting into the rate law of the RDS will produce a rate
law in terms of just reactants
85
An Example
2 NO(g)
k1
k-1
N2O2(g) Fast
H2(g) + N2O2(g)  H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]
H2(g) + N2O(g)  H2O(g) + N2(g) Fast
2 H2(g) + 2 NO(g)  2 H2O(g) + N2(g) Rateobs = k [H2][NO]2
for Step 1 Rateforward = Ratereverse
k1[NO]2 = k-1[N 2O2 ]
k1
[N 2O 2 ] =
[NO]2
k-1
Rate = k2 [H 2 ][N 2O 2 ]
k1
Rate = k2 [H 2 ] [NO 2 ]2
k-1
k2k1
Rate =
[H 2 ][NO2 ]2
k-1
86
Show that the proposed mechanism for the reaction 2 O3(g)
 3 O2(g) matches the observed rate law
Rate = k[O3]2[O2]-1
O3(g)
k1
k-1
O2(g) + O(g) Fast
O3(g) + O(g)  2 O2(g)
for Step 1 Rateforward = Ratereverse
k1[O3 ] = k-1[O 2 ][O]
k1
[O] =
[O3 ][O2 ]-1
k-1
Slow
Rate = k2[O3][O]
Rate = k2 [O3 ][O]
k1
Rate = k2 [O3 ] [O3 ][O 2 ]-1
k-1
k2k1
Rate =
[O3 ]2[O 2 ]-1
k-1
87
Catalysts
• catalysts are substances that affect the rate of a reaction without being
consumed
• catalysts work by providing an alternative mechanism for the reaction
 with a lower activation energy
• catalysts are consumed in an early mechanism step, then made in a
later step
mechanism without catalyst
mechanism with catalyst
O3(g) + O(g)  2 O2(g) V. Slow
Cl(g) + O3(g)  O2(g) + ClO(g)
Fast
ClO(g) + O(g)  O2(g) + Cl(g)
Slow
Tro, Chemistry: A Molecular Approach
88
Ozone Depletion over the
Antarctic
89
Energy Profile of a Catalyzed Reaction
polar stratospheric clouds contain
ice crystals that catalyze reactions
that release Cl from atmospheric
chemicals
90
Catalysts
• homogeneous catalysts are in the same
phase as the reactant particles
Cl(g) in the destruction of O3(g)
• heterogeneous catalysts are in a different
phase than the reactant particles
solid catalytic converter in a car’s exhaust
system
Tro, Chemistry: A Molecular Approach
91
Types of Catalysts
Tro, Chemistry: A Molecular Approach
92
Catalytic Hydrogenation
H2C=CH2 + H2 → CH3CH3
Tro, Chemistry: A Molecular Approach
93
Enzymes
• because many of the molecules are large
and complex, most biological reactions
require a catalyst to proceed at a
reasonable rate
• protein molecules that catalyze biological
reactions are called enzymes
• enzymes work by adsorbing the substrate
reactant onto an active site that orients it for
reaction
Tro, Chemistry: A Molecular Approach
94
Enzyme-Substrate Binding
Lock and Key Mechanism
Tro, Chemistry: A Molecular Approach
95
Enzymatic Hydrolysis of Sucrose
Tro, Chemistry: A Molecular Approach
96