Upload
ainul-aqila
View
297
Download
0
Embed Size (px)
Citation preview
7/27/2019 CHM096 1 Chem Kinetics Rm
1/100
7/27/2019 CHM096 1 Chem Kinetics Rm
2/100
1-2L.H. Sim
Kinetics: Rates and Mechanisms of Chemical Reactions
1.1 ExpressingReaction Rate: Differential rate law
Graphical representation of reaction rate
1.2 The Rate Law : Order of reaction, Initial rate method
Integrated rate law, rate constant, half -life
1.3 Collision Theory: Activation energy, effective collisions
1.6 Arrhenius Equation: Determination of rate constant and
activation energy
1.7 Reaction Mechanism: Intermediates, molecularity,
rate determining step
1.4 Factors affecting reaction rate: Concentration, temperature, catalyst,
nature of reactants
1.5 Transition State Theory: Activated complex
7/27/2019 CHM096 1 Chem Kinetics Rm
3/100
1-3L.H. Sim
1.1 Chemical Kinetics
The speed of a chemical reaction is called its
reaction rate
The rate of a reaction is a measure of how fast
the reaction makes products, or how fast the
reactants are consumed. Chemical kinetics is the study of the factors
that affect the rates of chemical reactions
Such as temperature, reactant
concentrations, addition of catalyst etc.
3
7/27/2019 CHM096 1 Chem Kinetics Rm
4/100
1-4L.H. Sim
Figure 1.1 Reaction rate: the central focus of chemical kinetics
7/27/2019 CHM096 1 Chem Kinetics Rm
5/100
1-5L.H. Sim
1.2 Expressing the Reaction Rate
Reaction rate is defined as the changes in the concentrations of
reactants or products per unit time
Reactant concentrations decrease while product concentrations
Increase. For the reactionA B
Where [ ] denote the concentration in mol dm-3 and , the change of
Unit for rate is mol dm-3 s-1 ormol/L s-1 or mol/dm3 min-1
Rate of reaction = concentration time
= - [A2] [A1]
t2-t1
= - [Reactant]
time
Or [product]Rate of reaction = +
t=
[B2] [B1]
t2-t
1
7/27/2019 CHM096 1 Chem Kinetics Rm
6/100
1-6L.H. Sim
Figure 1.2 The wide range of reaction rates.
7/27/2019 CHM096 1 Chem Kinetics Rm
7/1001-7L.H. Sim
06250061
87Rate
tt
XX
t
XRate
12
12
.
250061
84Rate
tt
AA
t
ARate
12
12
.
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
250061
04Rate
tt
CC
t
CRate
12
12
.
06250061
01Rate
tt
ZZ
t
ZRate
12
12
.
7
7/27/2019 CHM096 1 Chem Kinetics Rm
8/1001-8L.H. Sim
Initial rate is defined as the rate at the start of the reaction when an
infnitesimally small amount of the reactant has been used up and isgiven by the gradient to the curve at time t = 0
Instantaneous rate is the rate at a particular time and is given by the
tangent to the curve at that time in the concentration-time graph.
Average rate is the rate given by the average change in concentrationof a reactant or product per unit time over a certain time interval.
For
Average rate = [ B ] at time t2 [ B ] at time t1
t2 t1= [B]
t
t2 = final time ; t1 = initial time
1.3 Three different types of rates
A B
7/27/2019 CHM096 1 Chem Kinetics Rm
9/1001-9L.H. Sim
Table 1.1 Concentration of O3 at Various Time in its
Reaction with C2H4 at 303K
C2H4(g) + O3(g) C2H4O(g) + O2(g)
Time (s) Concentration of O3 (mol/L)
0.0
20.0
30.0
40.0
50.0
60.0
10.0
3.20x10-5
2.42x10-5
1.95x10-5
1.63x10-5
1.40x10-5
1.23x10-5
1.10x10-5
(conc O3)-t
7/27/2019 CHM096 1 Chem Kinetics Rm
10/1001-10L.H. Sim
Figure 1.3
The concentrations of O3 vs. time during its reaction with C2H4
- [C2H4]
trate =
[O2
]
t
or
(a) Initial rate, t = 0 s
(d) Instantaneous rate at t = 35 s
h
w
(b) Average rate from t = 10s to
t= 30s is concentration = htime w
rate =
C2H4(g) + O3(g) C2H4O(g) + O2 (g)
7/27/2019 CHM096 1 Chem Kinetics Rm
11/1001-11L.H. Sim
Figure 1.4 Plots of [C2H4] and [O2] vs. time.
7/27/2019 CHM096 1 Chem Kinetics Rm
12/1001-12L.H. Sim
In general, for the reaction
aA + bB cC + dD
rate =
1
a- = -
[A]t
1
b
[B]t
1
c
[C]t= +
1
d
[D]t= +
The numerical value of the rate depends upon the substance that
serves as the reference. The rest is relative to the balancedchemical equation.
1.4 Reaction Rate and Stoichiometry
7/27/2019 CHM096 1 Chem Kinetics Rm
13/1001-13
L.H. Sim
For the reaction below, the coefficients of the balanced equationare not all the same
H2 (g) + I2 (g) 2 HI(g)
Therefore, the change in the number of molecules of onesubstance is a multiple of the change in the number ofmolecules 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
To be consistent, the change in the concentration of eachsubstance is multiplied by 1/coefficient
13
7/27/2019 CHM096 1 Chem Kinetics Rm
14/1001-14
L.H. Sim
average rate in a given time
period = slope of the line
connecting the [H2] points; and
slope of the line for [HI]
the average rate for the
first 10 s is 0.0181 M/s
the average rate for the
first 40 s is 0.0150 M/s
the average rate for the
first 80 s is 0.0108 M/s
14
7/27/2019 CHM096 1 Chem Kinetics Rm
15/1001-15
L.H. Sim
Sample Problem 1.1
PLAN:
SOLUTION:
Expressing Rate in Terms of Changes in
Concentration with Time
PROBLEM: Because it has a nonpolluting product (water vapor), hydrogen
gas is used for fuel aboard the space shuttle and may be usedby Earth-bound engines in the near future.
2H2(g) + O2(g) 2H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O] increasing?
Choose [O2] as a point of reference since its coefficient is 1. For every
molecule of O2 which disappears, 2 molecules of H2 disappear and 2
molecules of H2O appear, so [O2] is disappearing at half the rate of change
of H2 and H2O.
- 12
[H2]t
[O2]
t
= +
[H2O]t
12
0.23mol/L*s= +
[H2O]
t
1
2; = 0.46mol/L*s[H2O]
t
rate =(a)
[O2]
t- = -(b)
= -
[O2]t
D t i R t f ti b d th
7/27/2019 CHM096 1 Chem Kinetics Rm
16/1001-16
L.H. Sim
Sample Problem 1.2Determine Rate of reaction based on the
stoichiometry of a reaction
PROBLEM: Dinitrogen pentoxide, N2O5, decomposes to form nitrogen dioxide
and oxygen.2 N2O5(g) 4NO2 (g) + O2(g)
NO2 is produced at a rate of 5.0 x 10-6 Ms-1. What is the
corresponding rate of disappearance of N2O5 and rate of
formation of O2?
(2.5 x 10-6, 1.25x 10-6M/s)
Sample Problem 1.3
PROBLEM: Ammonia,NH3, reacts with oxygen to form nitric oxide, NO and
water vapour.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)At a specific time in the reaction, NH3 is disappearing at a rate of
0.068 Ms-1. What is the corresponding rate of disappearance ofO2and production of NO and H2O.
(0.085 , 0.068 , 0.102 M/s)
7/27/2019 CHM096 1 Chem Kinetics Rm
17/1001-17
L.H. Sim
If 2.4 x 102 g of NOBr (MM 109.91 g) decomposes in a
2.0 x 102 mL flask in 5.0 minutes, find the average rate ofBr2 production in M/s.
2 NOBr(g) 2 NO(g) + Br2(l)
Sample Problem 1.4
M mole Br2 Rate min s }
0.018 M/s
7/27/2019 CHM096 1 Chem Kinetics Rm
18/1001-18
L.H. Sim
1.5 Rate Laws and order of reaction
A. Differential rate law or rate law is an experimentally
determined equation that expresses how the rate of a reaction
depends on the concentrations of the reactants
Example: 2N2O5(g) 4NO2(g) + O2(g)
Rate
[N2O5]x
= k [N2O5]x
where k = rate constant
x = order of reaction determined experimentally, not equivalent to the
stoichiometric coefficient of a balanced equation
7/27/2019 CHM096 1 Chem Kinetics Rm
19/100
1-19L.H. Sim
B. The order of reaction is defined as the power to which the
concentration of a reactant is raised in the rate equation or rate law.
It must be determined experimentally.
Overall orderof a reaction = sum of the powers of the concentration
terms in the rate law
The order of each reactant concentration explains how the rate of a
reaction varies with the concentration of a particular reactant. For
exampleRate = k [N2O5]
x, when [N2O5] is doubled,
If x = 0; the rate remains unchanged
If x = 1; the rate doubles
If x = 2; the rate quadruples / increase by a factor of 4
If x = 3; the rate increases eightfold / factor of 8
7/27/2019 CHM096 1 Chem Kinetics Rm
20/100
1-20L.H. Sim
For the reaction
A product
The proposed rate law : Rate = k[A]n
If a reaction is Zero Order, the rate of the reaction is
always the same
doubling [A] will have no effect on the reaction rate
If a reaction is First Order, the rate is directly
proportional to the reactant concentration
doubling [A] will double the rate of the reaction
If a reaction is Second Order, the rate is directlyproportional to the square of the reactant concentration
doubling [A] will quadruple the rate of the reaction
20
7/27/2019 CHM096 1 Chem Kinetics Rm
21/100
1-21L.H. Sim
For a general reaction, A B
Order ofreaction
Differential rate law Units of rate constant (k)(t in seconds)
zero Rate = k[A]o = k k = mol dm-3 s-1 or mol L s-1
The rate is independent of the
concentration of the reactant as
long as some is present.
First Rate = k[A] k = rate/[A]= mol dm-3 s-1
mol dm-3
= s-1
second Rate = k[A]2 k = rate/[A]2= mol dm-3 time-1
mol2 dm-6
= dm3mol-1 s-1 or L mol-1s-1
Table 1.2: order of reaction and differential rate law
7/27/2019 CHM096 1 Chem Kinetics Rm
22/100
1-22L.H. Sim
Sample Problem 1.5
SOLUTION:
Determining Reaction Order from Rate Laws
PROBLEM: For each of the following reactions, determine the reaction order
with respect to each reactant and the overall order from thegiven rate law.
(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2]
(b) CH3CHO(g) CH4(g) + CO(g); rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-
(aq) + 2H+
(aq) I3-
(aq) + 2H2O(l); rate = k[H2O2][I-
]
PLAN: Look at the rate law and not the coefficients of the chemical reaction.
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall.
(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,
while being 2nd order overall.
7/27/2019 CHM096 1 Chem Kinetics Rm
23/100
1-23L.H. Sim
In order to determine the order of a reaction, an
experiment has to be carried out to measure thereaction rates. The results obtained from these
experiments are used to determine the order of the
reaction.
The order of reaction and the rate constant can bedetermined by
a) the initial rate method
b) the half-life method
c) Unit of the rate constant
d) linear plots based on the integrated rate law
1.6 To determine the order of a reaction
7/27/2019 CHM096 1 Chem Kinetics Rm
24/100
1-24L.H. Sim
1.7 Experimental Determining of Reaction Orders
Initial rate method
Run a series of experiments, each of which starts with a different set
of reactant concentrations, and from each obtain an initial rate.
See Table 1.3 for data on the reaction
O2(g) + 2NO(g) 2NO2(g) rate = k[O2]m[NO]n
Compare 2 experiments in which the concentration of one reactant variesand the concentration of the other reactant(s) remains constant.
k[O2]2m[NO]2
n
k[O2]1m[NO]1
n=
rate2
rate1=
[O2]2m
[O2]1m
=
6.40x10-3mol/L*s
3.21x10-3mol/L*s
[O2]2
[O2]1
m
=1.10x10-2mol/L
2.20x10-2mol/L m
; 2 = 2m m = 1
Do a similar calculation for the other reactant(s).
7/27/2019 CHM096 1 Chem Kinetics Rm
25/100
1-25L.H. Sim
Table 1.3 Initial Rates for a Series of Experiments in the
Reaction Between O2 and NO
Experiment
Initial Reactant
Concentrations (mol/L)Initial Rate
(mol/L*s)
1
2
3
4
5
O2 NO
1.10x10-2
1.30x10-2
3.21x10-3
1.10x10-2 3.90x10-2 28.8x10-3
2.20x10-2
1.10x10-2
3.30x10-2
1.30x10-2
2.60x10-2
1.30x10-2
6.40x10-3
12.8x10-3
9.60x10-3
2NO(g) + O2(g) 2NO2(g)
7/27/2019 CHM096 1 Chem Kinetics Rm
26/100
1-26L.H. Sim
Sample Problem 1.6
PLAN:
SOLUTION:
Determining Reaction Order from Initial Rate Data
PROBLEM: Many gaseous reactions occur in a car engine and exhaust
system. One of these is
NO2(g) + CO(g) NO(g) + CO2(g) rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)
1
2
3
0.0050
0.080
0.0050
0.10
0.10
0.40
0.10
0.10
0.20
Solve for each reactant using the general rate law using the
method described previously.
rate = k[NO2]m[CO]n
First, choose two experiments in which [CO] remains
constant and the [NO2] varies.
7/27/2019 CHM096 1 Chem Kinetics Rm
27/100
1-27L.H. Sim
Sample Problem 1.6 Determining Reaction Order from Initial Rate Data
continued
0.080
0.0050
rate 2
rate 1
[NO2]2
[NO2]1
m=
k[NO2]m
2[CO]n
2
k[NO2]m
1 [CO]n
1
=
0.40
0.10=
m
;16 = 4m and m = 2
k[NO2]m
3[CO]n3
k[NO2]m
1 [CO]n1
[CO]3
[CO]1
n=
rate 3
rate 1=
0.0050
0.0050=
0.20
0.10
n;
1 = 2n and n = 0
The reaction is 2nd
order in NO2.
The reaction is
zero order in CO.
rate = k[NO2]2[CO]0 = k[NO2]
2
7/27/2019 CHM096 1 Chem Kinetics Rm
28/100
1-28L.H. Sim
Sample Problem 1.7
PROBLEM: Consider the following reaction and the data given in the
table, determine the rate law for the reaction and also
calculate the rate constant, k.
NH4+(aq) + NO2
-(aq) N2(g) + 2H2O(l)
Expt Initial [NH4+]
Initial [NO2-
] Initial rate (moldm-3
s-1)
1 0.10 0.005 1.35 x 10-7
2 0.10 0.01 2.70 x 10-7
3 0.20 0.01 5.40 x 10-7
7/27/2019 CHM096 1 Chem Kinetics Rm
29/100
1-29L.H. Sim
The following table shows the results for the reaction
BrO3-(aq) + 5Br-(aq) + 6H+(aq) 3Br2(l) + 3H2O(l)
Sample Problem 1.8
PROBLEM:
Expt Initial [BrO3- ] Initial [Br-] Initial [H+ ] Initial rate (mol dm-3s-1)
1 0.10 0.10 0.10 8.00 x 10-4
2 0.20 0.10 0.10 1.60 x 10-3
3 0.20 0.20 0.10 3.20 x 10-3
4 0.10 0.10 0.20 3.20 x 10-3
Determine the order of the reaction with respect to each
reactant and the value of the rate constant.
7/27/2019 CHM096 1 Chem Kinetics Rm
30/100
1-30L.H. Sim
1.8 Integrated Rate Laws
For the reaction A Products, the rate of the
reaction depends on the concentration of A Applying calculus to integrate the rate law gives
another equation showing the relationship between
the concentration of A and the time of the reaction
this is called the Integrated Rate Law
30
The integrated Rate Law expresses how the
concentrations of the reactants depend on time. The
expression of an integrated rate law depends on the
order of a reaction.
7/27/2019 CHM096 1 Chem Kinetics Rm
31/100
1-31L.H. Sim
Almost all experimental work in chemical kinetics deals
with integrated rate laws. The integrated rate laws
derived below are for simple reactions with only one
reactant.
(I) For zero-order reaction: A products
By the rate law (differential); Rate = - d[A] = k
dt
Integrating the differential rate law, we obtained the integrated
rate law
[A]t = -kt + [A ]o
where [A]o = initial concentration of A
A plot of[A]t versus t gives a straight line, slope = -k and
y-intercept = [A]o
7/27/2019 CHM096 1 Chem Kinetics Rm
32/100
1-32L.H. Sim
The half- life of a reaction, t1/2 is the time required for the concentration
of a reactant to decrease to half its original concentration
[A]o = -kt + [A]o
kt = [A]o
t = [A]o ; ** the half lifeof a zero orderreaction is
2k directly proportionalto [A]o
(II) For first- order reaction aA products
By the rate law , Rate = - d[A] = k[A]
dt
Integrated first-order rate law gives
ln[A]t = -kt + ln[A]o
ln [A]o = kt
[A]t
7/27/2019 CHM096 1 Chem Kinetics Rm
33/100
1-33L.H. Sim
A plot ofln [A]t versus t will give a straight line. slope =k
and y-intercept = ln [A]o.
The half-life is derived by rewriting the integrated rate law as
ln{ [A]o / [A]t } = kt
At t = t , [A]t = [A]o ; ln 2 = kt
0.693 = kt and t = 0.693/ k
** the half lifeof a first order reaction is independent of the initial
concentration of reactant. i.e 1st half-life (t 2nd half-life (t
(III) For second-order reaction aA products
By the rate law, Rate = -d[A] = k[A]2
dt
Integration of the rate law will give the integrated second order rate
law as
1 = kt + 1
[A]t [A]o
7/27/2019 CHM096 1 Chem Kinetics Rm
34/100
1-34L.H. Sim
At half-life , t , where [A]t = [A]o
1 = kt + 1
[A]o [A]o
kt = 2 - 1 = 1
[A]o [A]o [A]o
t = 1
k[A]o
A plot of1/[A]t versus t give a straight line with a slope = k
and the y-intercept = 1/[A]o
** The half life of a second order reaction is inversely proportional tothe initial concentration of the reactant
Thus , the second half-life (t ) = 2(t
7/27/2019 CHM096 1 Chem Kinetics Rm
35/100
1-35L.H. Sim
Summary of Integrated Rate Laws
rate = -[A]
t
= k[A]
rate = -[A]
t= k[A]0
rate = -
[A]
t= k[A]2
f ir st order rate equation
second order rate equation
zero order rate equati on
ln[A]0
[A]t
= kt ln [A]0 = kt + ln [A]t
1
[A]t
1
[A]0- = kt
1
[A]t
1
[A]0+= kt
[A]t - [A]0 = - kt
7/27/2019 CHM096 1 Chem Kinetics Rm
36/100
1-36L.H. Sim
1.8.1 Relationship Between Order and Half-Life
For a zero order reaction, the half-life is directlyproportional to the initial concentration. The lower theinitial concentration of the reactants, the shorter thehalf-life
t1/2= [A]init/2k
For a first order reaction, the half-life is independent ofthe concentration. (t1/2 = constant)
t1/2 = ln(2)/k
For a second order reaction, the half-life is inversely
proportional to the initial concentration increasing theinitial concentration shortens the half-life
t1/2 = 1/(k[A]init)
36
7/27/2019 CHM096 1 Chem Kinetics Rm
37/100
1-37L.H. Sim
Table 1.4 An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero OrderFirst Order Second Order
Plot for straight line
Slope, y-intercept
Half-life
Rate law rate = k rate = k[A] rate = k[A]2
Units fork mol/L*s S-1 L/mol*s
Integrated rate
law in straight-
line form
[A]t = -kt + [A]0 ln[A]t =
-kt + ln[A]0
1/[A]t =
kt + 1/[A]0
[A]t vs. t ln[A]t vs. t 1/[A]t vs t
-k, [A]0 -k, ln[A]0 k, 1/[A]0
[A]0/2k ln 2/k1/k[A]0
7/27/2019 CHM096 1 Chem Kinetics Rm
38/100
1-38L.H. Sim
Figure 1.5 Integrated rate laws and reaction order
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
[A]t = -kt + [A]0
f O f f
7/27/2019 CHM096 1 Chem Kinetics Rm
39/100
1-39L.H. Sim
Figure 1.6 A plot of [N2O5] vs. time for three half-lives.
For a first-order process, the half-life is a
constant and is independent of initial
concentration ln 2 0.693T =
k k=
7/27/2019 CHM096 1 Chem Kinetics Rm
40/100
1-40L.H. Sim
Figure 1.7 Graphical determination of the reaction order for the
decomposition of N2O5.
The decomposition
reaction of N2O5 is afirst order reaction,
only the plot of ln[N2O5]
vs t is linear. Plots of
[N2O5] vs t and
1/[N2O5] vs t not linear.
7/27/2019 CHM096 1 Chem Kinetics Rm
41/100
1-41L.H. Sim
Sample Problem 1.9
PLAN:
SOLUTION:
Determining Reaction Concentration at a Given Time
apply integrated rate law
PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order
reaction, with the very high rate constant of 87s-1, to twomolecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is the
concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?
Find the [C4H8] at time, t, using the integrated rate law for a 1st order
reaction. Once that value is found, divide the amount decomposed by
the initial concentration.
; ln2.00
[C4H8]t
= (87s-1)(0.010s)
[C4H8]t = 0.83mol/L
ln[C4H8]0
[C4H8]t
= kt
(a)
(b) [C4H8]0 - [C4H8]t
[C4H8]0
=2.00M - 0.87M
2.00M= 0.58
7/27/2019 CHM096 1 Chem Kinetics Rm
42/100
1-42L.H. Sim
Sample Problem 1.10
PROBLEM: The decomposition of N2O5 is a first-order reaction with a rate
constant of 5.1 x 10-4
s-1
at 45o
C.2N2O5 (g) 4NO (g) + O2(g)
a) If the initial concentration of N2O5 is 0.25 M, what is
the concentration after 3.2 min?
b) How long will it take for the concentration of N2O5 to
decrease from 0.25M to 0.15M?
c) How long will it take to convert 62% of the starting
material?
d) Calculate the half-life of the reaction .
(0.25 M, 17 min, 32 min)
7/27/2019 CHM096 1 Chem Kinetics Rm
43/100
1-43L.H. Sim
Sample Problem 1.11
PROBLEM: Butadiene reacts to form its dimer according to the equation
2C4H6(g) C8H12(g)The following data were collected for this reaction at a given
temperature.
[C4H6(g) ] moldm-3 Time (s) [C4H6(g) ] moldm
-3 Time (s)
0.01000 0.000 0.00313 36000.00625 1000 0.00270 4400
0.00476 1800 0.00241 5200
0.00370 2800 0.00208 6200
a) Is the reaction first order or second order?
b) What is the value of the rate constant for the reaction?
c) What is the half-life of the reaction?
7/27/2019 CHM096 1 Chem Kinetics Rm
44/100
1-44 L.H. Sim
Sample Problem 1.12
PLAN:
SOLUTION:
Determining the Half-Life of a First-Order Reaction
PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its
600 bond angles allow poor orbital overlap, its bonds are weak.As a result, it is thermally unstable and rearranges to propene at
10000C via the following first-order reaction:
CH2
H2C CH2(g)
H3C CH CH2 (g)
The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How
long does it take for the concentration of cyclopropane to reach one-quarter of
the initial value?
Use the half-life equation, t =0.693
k, to find the half-life.
One-quarter of the initial value means two half-lives have passed.
t = 0.693/9.2s-1 = 0.075s(a) 2 t = 2(0.075s) = 0.150s(b)
S l P bl 1 13
7/27/2019 CHM096 1 Chem Kinetics Rm
45/100
1-45 L.H. Sim
The reaction Q 2 R is second order in Q. If the initial
[Q] = 0.010 M and after 5.0 x 102
seconds the [Q] =0.0010 M,
(i) find the rate constant.
(ii) what is the length of time for [Q] = [Q]initial
45
Sample Problem 1.13
7/27/2019 CHM096 1 Chem Kinetics Rm
46/100
1-46 L.H. Sim
(ii)
(i)
7/27/2019 CHM096 1 Chem Kinetics Rm
47/100
1-47 L.H. Sim
1.9 Collision Theory: Basis of the Rate Law
The Collision theoryis developed from the kinetic theory to explain
the effects of concentration and temperature on reaction rates.
The basic concept of collision is that all chemical reactions occur as a
result of collisions between reacting molecules. However, for a
reaction between two particles to occur, an effective collision must
take place.
Two requirements must be satisfied foreffective collisions betweentwo reacting molecules to take place.
1. The reacting molecules must collide with energy which must be
equal to or greater than the activation energy of the reaction (i.e.
the minimum amount of energy required to initiate a chemical
reaction). (Fig 1.8 & 1.9)
2. The molecules must approach each other in the right orientation
before a collision could lead to product formation. (Fig 1.11)
1 9 1 Activation Energy
7/27/2019 CHM096 1 Chem Kinetics Rm
48/100
1-48 L.H. Sim
1.9.1 Activation Energy
The Activation Energyis defined as the minimum amount of kinetic
energy required for a chemical reaction to take place and form
products.This energy is needed to break chemical bonds of the reactants and to
rearrange atoms and valence electrons to form the products as the
reaction proceeds.
Reaction with low activation energy is easier to take place, and is faster
A reaction profile or energy profile is a plot of energies (potential
energy) versus progress of reaction as shown in Fig 1.9 & 1.14. The
progress of reaction represents the extent of the reaction.
Fig 1.10 (a) and (b) shows the reaction profiles of an exothermic and
endothermic reactions, respectively.
Ea = Activation energy
H = Enthalpy of reaction or Heat of reaction
7/27/2019 CHM096 1 Chem Kinetics Rm
49/100
1-49 L.H. Sim
1.9.2 Effective Collisions depend on TWO Factors
For a collision to
be effective, the
reacting molecules
must havesufficient kinetic
energy to
overcome the
energy barrier.
49
Figure 1.8I. Kinetic Energy Factor
Fig re 1 9
7/27/2019 CHM096 1 Chem Kinetics Rm
50/100
1-50 L.H. Sim50
Figure 1.9
7/27/2019 CHM096 1 Chem Kinetics Rm
51/100
1-51 L.H. Sim
heat inH > 0
Ea
Figure 1.10 Reaction profiles of an exothermic and endothermic
reaction
H < 0H < 0H < 0H < 0H < 0H < 0H < 0H < 0
heat outH < 0
Ea
(a) Exothermic reaction (b) Endothermic reaction
7/27/2019 CHM096 1 Chem Kinetics Rm
52/100
1-52 L.H. Sim
II. Orientation Factor
The proper orientation results when the atomsare aligned in such a way that the old bonds canbreak and the new bonds can form
The more complex the reactant molecules, theless frequently they will collide with the properorientation
52
Fi 1 11
7/27/2019 CHM096 1 Chem Kinetics Rm
53/100
1-53 L.H. Sim53
Figure 1.11
ON
ClO
O
O
Cl
Cl
Cl
7/27/2019 CHM096 1 Chem Kinetics Rm
54/100
1-54 L.H. Sim
1.10 The Transition State Theory - Molecular nature
of the Activated complex
There is an energy barrier (activation energy) to almost allreactions
There exists an intermediate stage when reactant molecules
change into products.
The activated complex ortransition state is a chemical
species with partially broken and partially formed bonds
an intermediate formed between reactant and product
molecules
has highest energy and extremely unstable
54
Ea(forw) = activation energy for the forward reaction(from reactant to peak)
Ea(rev) = activation energy for the reverse reaction
(from product to peak)
7/27/2019 CHM096 1 Chem Kinetics Rm
55/100
1-55 L.H. Sim
Figure 1.12 Energy-level diagram for a reaction
A reaction can occur in either direction, so the diagram shows two
activation energies. The forward reaction is exothermic because
the reactants have more energy than the products and the
Ea(forward) < Ea(reverse)
Collision
Energy
ACTIVATEDSTATE
Ea
(reverse)
PRODUCTS
REACTANTS
Ea
(forward)
CollisionEnergy
For molecules to
react, they must
collide with enough
energy to reach an
activated state. Thisminimum collision
energy is the
activation energy,
Ea
A l l di f th f ti f lli i
7/27/2019 CHM096 1 Chem Kinetics Rm
56/100
1-56 L.H. Sim
Figure 1.13 An energy-level diagram of the fraction of collisions
exceeding Ea.
R ti fil f th ti f CH B d OH
7/27/2019 CHM096 1 Chem Kinetics Rm
57/100
1-57 L.H. Sim
Figure 1.14 Reaction energy profile for the reaction of CH3Br and OH-.
1 2 3 4 5
R f t th ti fil f th ti b t CH B
7/27/2019 CHM096 1 Chem Kinetics Rm
58/100
1-58 L.H. Sim
Refer to the reaction energy profile for the reaction between CH3Br
and OH- shown in Fig 1.14:
CH3Br + OH- CH3OH + Br
[1] Two molecules that are far apart but speeding toward each otherhave high kinetic energy and low potential energy. As the
molecules get closer some kinetic energy is converted to potential
energy as the electron clouds repel each other.
[2] When the molecules collide, kinetic energy is changed into potentialenergy because the molecules are distorted during collision to
break bonds and rearrange the atoms ready to form product
molecules. If this potential energy is less than the activation energy
the molecules bounce off without forming product.
[3] A fraction of the molecules which collide with sufficient kineticenergy along the line of approach to carry them over the
activation barrier, the peak in the reaction profile, then reaction
occur. At the peak, the C-Br bond lengthen and weaken while the
C-OH bond began to form.
7/27/2019 CHM096 1 Chem Kinetics Rm
59/100
1-59 L.H. Sim
At this point during the transformation, the two reacting molecules are
highly distorted, forming the highly energetic and unstable species called
the transition state, oractivated complex as shown in Fig 1.14 (3)
and Fig 1.15. The dotted line in the Fig represents bonds in the processof breaking and making. C is surrounded by 5 atoms
This complex only forms if the molecules collide in an effective orientation
and with energy equal to or greater than the activation energy. Thus the
Ea is the quantity needed to stretch and deform bonds in order to reach
the transition state
[4] The potential energy of the complex starts to decrease as the new
bond C-O bond in the product shorten and strengthen and the C-Br
bond in the reactant molecule weakens.
[5] At separations to the right of the maximum, the potential energyrapidly falls to a low value as the product molecules separate .
7/27/2019 CHM096 1 Chem Kinetics Rm
60/100
1-60 L.H. Sim
Figure 1.15
Nature of the transition state in the reaction between CH3
Br and OH-.
CH3Br + OH- CH3OH + Br
-
transition state or activated complex
Figure 1 16
7/27/2019 CHM096 1 Chem Kinetics Rm
61/100
1-61 L.H. Sim
Figure 1.16
Reaction energy diagrams and possible transition states.
7/27/2019 CHM096 1 Chem Kinetics Rm
62/100
1-62 L.H. SimReaction progress
Potential
Energy
Sample Problem 1.14
SOLUTION:
Drawing Reaction Energy profile
PROBLEM: A key reaction in the upper atmosphere is
O3(g) + O(g) 2O2(g)
The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a
reaction energy diagram for this reaction, postulate a transition state, and
calculate Ea(rev).
PLAN: Consider the relationships among the reactants, products and
transition state. The reactants are at a higher energy level than theproducts and the transition state is slightly higher than the
reactants.
O3+O
2O2
Ea= 19kJ
Hrxn = -392kJ
Ea(rev)= (392 + 19)kJ =
411kJ
OO
O
O
breakingbond
formingbond
transition state
1 11 Factors Affecting Reaction Rate
7/27/2019 CHM096 1 Chem Kinetics Rm
63/100
1-63 L.H. Sim
1.11 Factors Affecting Reaction Rate
Any factor that can increase the frequency of effective
collisions would increase the rate of a reaction
Under specific conditions, every reaction has its own
characteristic rate, which may be controlled by four factors:
1. Concentrations of reactants
2. Chemical nature and physical state of reactants
3. Temperature
4. The use of a catalyst
Tro: Chemistry: A Molecular Approach 63
(A) Concentration of reactants
7/27/2019 CHM096 1 Chem Kinetics Rm
64/100
1-64 L.H. Sim
(A) Concentration of reactants
Rate of reaction number of collision
Second
OR Rate of reaction frequency of collision
According to the collision theory, as the number of reacting
molecules in a given volume increase, the frequency of collisions
(the number of collisions that happen per unit time) will be increasewhich then increase the amount ofeffective collisions (the
probability of collisions with sufficient energies for a reaction to
occur). This lead to an increase in the rate of a reaction. (Fig 1.17)
Concentration of gases depends on the partial pressure of the gas
higher pressure = higher concentration
Concentrations of solutions depend on the solute-to-solution ratio
(molarity)
Fi 1 17
7/27/2019 CHM096 1 Chem Kinetics Rm
65/100
1-65 L.H. Sim
Figure 1.17
The dependence of possible collisions on the product
of reactant concentrations.
A
A
B
B
A
A
B
B
A
4 collisions
Add another
molecule of A6 collisions
Add another
molecule of B
A
A
B
B
A B
(B) Nature of the Reactants
7/27/2019 CHM096 1 Chem Kinetics Rm
66/100
1-66 L.H. Sim
(B) Nature of the Reactants
Nature of the reactants means what kind of reactant molecules
and what physical state 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 leading tohigher frequency of collision. The probability of effective
collision will be increased. Surface area can be increased by
grinding the solids into fine powders. (Fig 1.18)
certain types of chemicals are more reactive than others
e.g. potassium metal is more reactive than sodium
ions react faster than molecules
no bonds need to be broken
66
7/27/2019 CHM096 1 Chem Kinetics Rm
67/100
1-67 L.H. Sim
Figure 1.18 The effect of surface area on reaction rate.
(C) Temperature
7/27/2019 CHM096 1 Chem Kinetics Rm
68/100
1-68 L.H. Sim
Increasing the temperature raises the average kinetic energy of the
reactant molecules.
Increasing the temperature increases the number of molecules with
sufficient kinetic energy to overcome the activation energy, leads to
an increase in the frequency of effective collisions between
molecules. (Fig 1.19)
The distribution of kinetic energy is shown by the Boltzmanndistribution curves shown in Fig 1.19
Fig 1.19 shows that at both temperatures T1 and T2 (T2 .> T1), a
relatively small fraction of collisions have sufficient kinetic energy
the activation energy to result in a reaction. Moreover, the
fraction ofeffective collisionsor collisions with the requiredactivation energy increases exponentially with an increasein
temperature from T1 to T2 (see Fig 1.20).
( ) p
Figure 1 19
7/27/2019 CHM096 1 Chem Kinetics Rm
69/100
1-69 L.H. Sim
Figure 1.19
The effect of temperature on the distribution of collision energies
1 12 The Arrhenius Equation The Effect of Temperature
7/27/2019 CHM096 1 Chem Kinetics Rm
70/100
1-70 L.H. Sim
1.12 The Arrhenius Equation The Effect of Temperature
on Reaction Rate
k AeEaRT
where kis the kinetic rate constant at T
Ea is the activation energy
R is the gas constant = 8.314 J K-1mol-1
T is the Kelvin temperature
A is the collision frequency factor
The exponential factor in the Arrhenius
equation is a number between 0 and 1
The larger the activation energy, Ea, the fewer molecules that have
sufficient energy to overcome the energy barrier, the smaller the rate
constant k.
Increase in temperature, T, will gives a largerkvalue therefore
increasing the temperature will increase the reaction rate
This type of relationship is called an exponential relationship
7/27/2019 CHM096 1 Chem Kinetics Rm
71/100
1-71 L.H. Sim71
yp p p p
which means that a small rise in temperature results in a large
increase in the fraction of molecules collide with energy equal to
or greater than the activation energy (Fig 1.19).
1.13 Arrhenius Plots
The Arrhenius Equation: The integrated form
This equation is in the linear form y= mx+ b
where y= ln(k) andx= (1/T)
A graph ofln(k) vs. (1/T) is a straight line
eyintercept =A (unit is the same as k)
7/27/2019 CHM096 1 Chem Kinetics Rm
72/100
1-72 L.H. Sim
Figure 1.20 Dependence of the rate constant on temperature
7/27/2019 CHM096 1 Chem Kinetics Rm
73/100
1-73 L.H. Sim
Figure 1.21 Graphical determination of the activation energy
ln k= -Ea/R (1/T) + ln A
Sample Problem 1.15
7/27/2019 CHM096 1 Chem Kinetics Rm
74/100
1-74 L.H. Sim
Determine the activation energy and frequency factor for the reaction
O3(g) O2(g) + O(g) given the following data:
74
Plot the graph ln kversus 1/T. From the straight line curve,
7/27/2019 CHM096 1 Chem Kinetics Rm
75/100
1-75 L.H. Sim75
slope = - Ea/R
y-intercept = ln (frequency factor, A). Figure 1.22
slope m = -1 12 x 104 K
7/27/2019 CHM096 1 Chem Kinetics Rm
76/100
1-76 L.H. Sim76
slope, m 1.12 x 10 K
yintercept, b = 26.8
Or ln A = 26.8, anti ln (26.8) = 4.36 x 1011 M-1.s-1
7/27/2019 CHM096 1 Chem Kinetics Rm
77/100
1-77 L.H. Sim
Table 1.5 The Effect of Ea and T on the Fraction (f) of Collisions
with Sufficient Energy to Allow Reaction
Ea (kJ/mol) f(at T = 298 K)
50 1.70x10-9
75 7.03x10-14
100 2.90x10-18
T f(at Ea = 50 kJ/mol)
250C(298K) 1.70x10-9
350C(308K) 3.29x10-9
450C(318K) 6.12x10-9
1 14 A h i E ti T P i t F
7/27/2019 CHM096 1 Chem Kinetics Rm
78/100
1-78 L.H. Sim
1.14 Arrhenius Equation: Two-Point Form
If only two data points i.e two T and two kare
given, the following forms of the Arrhenius
Equation can be used:
78
Sample Problem 1.16 Determining the Energy of Activation
7/27/2019 CHM096 1 Chem Kinetics Rm
79/100
1-79 L.H. Sim
PLAN:
SOLUTION:
PROBLEM: The decomposition of hydrogen iodide,
2HI(g) H2(g) + I2(g)
has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at
600. K. Find Ea.
Use the modification of the Arrhenius equation to find Ea.
ln
k2
k1 =
Ea
R
1
T1
1
T2
-
R= - (8.314J/mol*K)
Ea = 1.76x105 J/mol = 176 kJ/mol
ln1
500K
1
600K-
1.10x10-5L/mol*s
9..51x10-9L/mol*s=
8.314 J/Mol*K
Ea
100500 x 600
ln (0.11567 x 104) =8.314 J/Mol*K
Ea
7.0533 Ea x 4.009 x 10-5
=
Sample Problem 1.17
7/27/2019 CHM096 1 Chem Kinetics Rm
80/100
1-80 L.H. Sim
Determining ActivationEnergy graphically
PROBLEM:
The 2nd order rate constant for the decomposition of N2O
into N2 and O2 has been measured at a series of
temperatures.
(a) Determine graphically the activation energy for the
reaction.
(b) Refer to the graph, predict the rate constant at 800K
k (1/M) 1.87x 10-3
0.0113 0.0569 0.244
t (oC) 600 650 700 750
Figure 1.23
7/27/2019 CHM096 1 Chem Kinetics Rm
81/100
1-81 L.H. Sim
Information sequence to determine the kinetic parameters of a reaction.
Series of plots
of concentra-
tion vs. time Initialrates Reaction
ordersRate constant
(k) and actual
rate law
Integrated
rate law(half-life,
t1/2)
Rate constant
and reaction
order
Activation
energy, Ea
Plots of
concentration
vs. time
Find k atvaried T
Determine slope
of tangent at t0 for
each plot
Compare initial
rates when [A]
changes and [B] isheld constant and
vice versa
Substitute initial rates,
orders, and concentrationsinto general rate law:
rate = k[A]m[B]n
Use direct, ln or
inverse plot to
find order
Rearrange to
linear form and
graph
Find k atvaried T
1.15 REACTION MECHANISMS
7/27/2019 CHM096 1 Chem Kinetics Rm
82/100
1-82 L.H. Sim
The reaction mechanismof a chemical reaction is the sequence of
reaction steps orelementary reaction sthat ultimately lead from the
initial reactants to the final products of a reaction.
Anelementary reaction or elementary steprepresents a single
molecular event, such as dissociation or collision of molecules, in the
progress of the overall reaction.
An Example of a Reaction Mechanism
Overall reaction:
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
Plausible Mechanism:
Step 1: H2(g) + ICl(g) HCl(g) + HI(g)
Step 2: HI(g) + ICl(g)
HCl(g) + I2(g) The reactions (1) and (2) in this mechanism are called the
elementary steps, meaning that they cannot be broken downinto simpler steps, and that the molecules actually interact directlyin this manner without any other steps
7/27/2019 CHM096 1 Chem Kinetics Rm
83/100
1-83 L.H. Sim
A catalystdoes not appear in the net equation, it is consumed in the
first step and produced in the second or last step of the reaction
mechanism.
Areaction intermediate is a substance produced during a
elementary reaction that does not appear in the net equation because
it reacts in a subsequent step or elementary reaction in the
mechanism. In other words, Reactionintermediates are products inan early mechanism step, but then a reactant in a later step
E.g: Overall reaction: 2H2O2(aq) 2H2O(l) + O2(g)
Plausible mechanism : H2O2 + I- H2O + OI
-
H2O2 + OI- H2O + O2 + I
-
OI- = intermediate; I- = catalyst
1.15.1 Rate Laws for elementary steps
7/27/2019 CHM096 1 Chem Kinetics Rm
84/100
1-84 L.H. Sim
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
directly from the equation of the step
Overall: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)
Elementary steps in the reaction mechanism:
1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl]
2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]
84
7/27/2019 CHM096 1 Chem Kinetics Rm
85/100
1-85 L.H. Sim
Molecularity refers to the number of species (free atoms, ions or
molecules) that must collide or dissociates to produce the reaction
indicated by a particular elementary step.
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
Molecularity
Table 1.6 Rate Laws for General Elementary Steps
Rate Law
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = k[A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
A product
2A product
Elementary Step
1.15.2 The Rate-Determining Step of a Reaction Mechanism
7/27/2019 CHM096 1 Chem Kinetics Rm
86/100
1-86 L.H. Sim
The overall rate of a reaction is related to the rate of the
slowest, or rate-determining step. The slowest step or rate
determining step has the largest activation energy
The rate law of a reaction is determined by the rate of the rate -
determining step of the acceptable mechanism of the reaction.
The rate law of the rate determining step determines the rate
law of the overall reaction.
1.15.3 A plausible reaction Mechanism must satisfy two
requirements
a) The elementary steps must be physically reasonable and add
up to the overall equation.
b) The mechanism must correlated with the experimentally
determined rate law.
1.15.4 A mechanism with a slow step followed by a fast step
7/27/2019 CHM096 1 Chem Kinetics Rm
87/100
1-87 L.H. Sim
A mechanism can never be proved absolutely correct. We can only
say that a mechanism which satisfy the two requirements is
possibly correct or acceptable .
Example :
The balanced equation for the reaction of NO2(g) and F2(g) is
2NO2(g) + F2(g) 2NO2F(g)
The experimentally rate law is Rate = k[NO2][F2]
A suggested mechanism for the reaction is
k1
NO2 + F2 NO2F + F slowk2F + NO2 NO2F fast
Is this an acceptable mechanism for the reaction?
Requirement 1: Add up the two elementary reactions
7/27/2019 CHM096 1 Chem Kinetics Rm
88/100
1-88 L.H. Sim
Requirement 1: Add up the two elementary reactions
k1NO2 + F2 NO2F + F slow
k2F + NO2 NO2F fast
2NO2(g) + F2(g) 2NO2F(g) (same as the overall reaction
given)
Requirement 2: write the rate law based on the rate determining step
(i.e. the slow step)
Rate = k1[NO2][F2] (same as the experimentally determined
rate law given )
The mechanism is acceptable as it satisfies the 2 requirements.
Another Reaction Mechanism
7/27/2019 CHM096 1 Chem Kinetics Rm
89/100
1-89 L.H. Sim
The first step in this mechanism
is the rate determining step.
The first step is slower than the
second step because its
activation energy is larger.
The rate law of the first step is
the same as the rate law of theoverall reaction.
89
NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2
1. NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 Slow
2. NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast
Sample Problem 1.18
Determining Molecularity and Rate Laws for
7/27/2019 CHM096 1 Chem Kinetics Rm
90/100
1-90 L.H. Sim
PLAN:
SOLUTION:
Determining Molecularity and Rate Laws for
Elementary Steps
PROBLEM: The following two reactions are proposed as elementary steps in
the mechanism of an overall reaction:(1) NO2Cl(g) NO2(g) + Cl(g)
(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(a) The overall equation is the sum of the steps.
(b) The molecularity is the sum of the reactant particles in the step.
2NO2Cl(g) 2NO2(g) + Cl2(g)
(c) Write the rate law for each step.
rate2 = k2 [NO2Cl][Cl]
(1) NO2Cl(g) NO2(g) + Cl(g)
(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)
(a)
Step(1) is unimolecular.
Step(2) is bimolecular.
(b)
rate1 = k1 [NO2Cl](c)
Sample Problem 1.19
7/27/2019 CHM096 1 Chem Kinetics Rm
91/100
1-91 L.H. Sim
PROBLEM: (a) Consider the following two-step reaction mechanism for a
chemical reaction.
Step 1. NO2(g) + NO2(g) NO3(g) + NO(g) ..(slow)
Step 2. NO3(g) + CO(g) NO2(g) + CO2(g).(fast)
(i) Write the chemical equation for the overall reaction.
(ii) Identify the reaction intermediate
(iii) What is the order of the reaction?
(b) The decomposition of hydrogen iodide occurs via the action of the
following elementary steps
I. HI + HI H2I2 .Slow
II. H2I2 H2 + 2I fast
III. 2I I2 .fast
(i) Write the rate law of the reaction and state the order of thereaction
(ii) What is the molecularity of each of the elementary step?
(iii) Write the stoichiometric equation for the overall reaction.
(iv) State the reaction intermediate.
1.15.5 Mechanism which involve a fast reversible step
7/27/2019 CHM096 1 Chem Kinetics Rm
92/100
1-92 L.H. Sim
Example: The reaction of Cl2 with CHCl3 is given by the equation
Cl2(g) + CHCl3(g) HCl(g) + CCl4(g)
The rate law determined experimentally is Rate = k[Cl2] [CHCl3]
A proposed mechanism for the reaction is
k1Cl2 2Cl fast
k -1
k2Cl + CHCl3 HCl + CCl3 slow
k3
CCl3 + Cl CCl4 fast
Is this an acceptable mechanism for the reaction?
I) Cl2 2ClCl + CHCl3 HCl + CCl3
CCl3 + Cl CCl4
Cl2(g) + CHCl3(g) HCl(g) + CCl4(g)
II) Rate = k2[Cl][CHCl3]
Cl i i t di t hi h i ll d t b i th t l T b tit t
7/27/2019 CHM096 1 Chem Kinetics Rm
93/100
1-93 L.H. Sim
Cl is an intermediate which is allowed to be in the rate law. To substitute
the intermediate, we make use of the equilibrium of the fast step.
At equilibrium, rate of forward reaction=rate of reverse reaction
k1[Cl2] = k-1[Cl]2
[Cl] = (k1/k-1) [Cl2]
Substitute into the above rate law, Rate = k2(k1/k-1)[Cl2]
[CHCl3]
= k[Cl2][CHCl3]
Since the mechanism satisfy the two requirements,
it is acceptable
Sample Problem 1.20
PROBLEM: Consider the following reaction
2O3(g) 3O2(g)
The experimentally rate law is Rate = k [O3]2
[O2]
The mechanism proposed for the reaction is
k1
O O + O ( fast equilibrium)
7/27/2019 CHM096 1 Chem Kinetics Rm
94/100
1-94 L.H. Sim
O3 O2 + O ( fast equilibrium)
k -1
k2
O + O3 2O2 (slow)Verify the acceptability of the proposed mechanism for the above reaction.
Sample Problem 1.21
The equation for the reaction is Tl3+
+ Hg22+
Tl+
+ 2Hg2+
The experimental determined rate law is : rate = k [Tl3+] [Hg22+]
[Hg2+]
Verify the acceptability of the proposed mechanisms below :
k1
Hg22+ Hgo + Hg2+ (fast)k -1
k2Tl3+ + Hgo Tl+ + Hg2+ (slow)
1.16 CATALYSTS
7/27/2019 CHM096 1 Chem Kinetics Rm
95/100
1-95 L.H. Sim
Acatalystis a substance that increases the rate of a chemical
reaction without itself being consumed.
In general a catalyst lowers the activation energy, Ea, of the reaction
by providing a different mechanism or energy pathway for the
reaction concerned. Lowering Ea increases the rate constant, k, and
thereby increases the rate of the reaction
A catalyst increases the rate of the forwardand the reverse
reactions to the same extent.
A catalyzed reaction yields the products more quickly, but does not
yield more product than the uncatalyzed reaction.
Two types of catalysis:
a) Homogenous catalyst is one that is present in the same phaseor physical state as the reactant molecules. Eg enzyme reaction
b) Heterogeneous catalyst is in a different phase from the
reacting molecules, usually in the solid phase .
Figure 1.24
7/27/2019 CHM096 1 Chem Kinetics Rm
96/100
1-96 L.H. Sim
Reaction energy profile of a catalyzed and an uncatalyzed process.
Examples of reactions catalyzed by homogeneous catalysts
7/27/2019 CHM096 1 Chem Kinetics Rm
97/100
1-97 L.H. Sim
1. The depletion of ozone in the stratosphere by Cl atoms is an
example ofhomogeneous catalysis.
Cl(g) + O3(g) ClO(g) + O2(g)ClO(g) + O(g) Cl(g) + O2(g)
O3(g) + O(g) 2O2(g) ( Cl catalyst)
The uncatalyzed reaction is O3 + O 2O2 .
2. The decomposition of H2O2 catalyzed by I- ion.H2O2 + I
- H2O + IO-
H2O2 + IO- H2O + O2 + I
-
The uncatalyzed reaction is 2H2O2 2H2O + O2 .
3. The hydrolysis of an organic ester (RCOOR) to form a carboxylicacid (RCOOH), the reaction is catalyzed by a strong acid (H+).
R C O R + H2O R C OH + R OH (uncatalyzed)
OO
Figure 1.25 The metal-catalyzed hydrogenation of ethylene-
7/27/2019 CHM096 1 Chem Kinetics Rm
98/100
1-98 L.H. Sim
g e eta cata y ed yd oge at o o et y e e
heterogeneous catalysis
H2C CH2 (g) + H2 (g) H3C CH3 (g)Ni
7/27/2019 CHM096 1 Chem Kinetics Rm
99/100
1-99 L.H. Sim
CH3CH CH2 + H2 CH3CH CH2
HH
Ni
Example 2: Hydrogenation of an alkene catalyzed by Ni
propenepropane
T bl 1 7
7/27/2019 CHM096 1 Chem Kinetics Rm
100/100
Table 1.7