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16-1
Chapter 4
Kinetics: Rates and Mechanisms
of Chemical Reactions
16-2
Kinetics: Rates and Mechanisms of Chemical Reactions
16.1 Factors That Influence Reaction Rate
16.2 Expressing the Reaction Rate
16.3 The Rate Law and Its Components
16.4 Integrated Rate Laws: Concentration Changes over Time
16.7 Reaction Mechanisms: Steps in the Overall Reaction
16.8 Catalysis: Speeding Up a Chemical Reaction
16.5 The Effect of Temperature on Reaction Rate
16.6 Explaining the Effects of Concentration and Temperature
16-3
Figure 16.1 Reaction rate: the central focus of chemical kinetics.
16-4
Figure 16.2 The wide range of reaction rates.
16-5
Factors That Influence Reaction Rate
Under a specific set of conditions, every reaction has its own
characteristic rate, which depends upon the chemical nature of
the reactants.
Four factors can be controlled during the reaction:
1. Concentration - molecules must collide to react;
2. Physical state - molecules must mix to collide;
3. Temperature - molecules must collide with enough energy to react;
4. The use of a catalyst.
16-6
Figure 16.3 The effect of surface area on reaction rate.
16-7
Figure 16.4 Collision energy and reaction rate.
16-8
Expressing the Reaction Rate
reaction rate - changes in the concentrations of reactants or
products per unit time
reactant concentrations decrease while product concentrations
increase
Rate of reaction = -
for A B
change in concentration of A
change in time
= -conc A2 - conc A1
t2 - t1
(conc A)-
t
16-9
Table 16.1 Concentration of O3 at Various Times in its
Reaction with C2H4 at 303 K
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 A)-
t
16-10
Figure 16.5
The concentration of O3 vs. time during its reaction with C2H4.
C2H4(g) + O3(g) C2H4O(g) + O2(g)
- [C2H4]
t
rate =
- [O3]
t
=
16-11
Figure 16.6 Plots of [C2H4] and [O2] vs. time.
Tools of the
Laboratory
16-12
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 balanced
chemical equation.
16-13
Sample Problem 16.1
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 used
by earthbound 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.
-1
2
[H2]
t= -
[O2]
t= +
[H2O]
t
1
2
0.23 mol/L*s = +[H2O]
t
1
2= 0.46 mol/L*s
[H2O]
t
rate =(a)
[O2]
t- = -(b)
16-14
Sample Problem 16.2
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 the
given 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.
16-15
Table 16.2 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)
16-16
Determining Reaction Orders
Using initial rates -
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 16.2 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
varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2
n
k [O2]1m[NO]1
n=
rate 2
rate 1 =
[O2]2m
[O2]1m
=
6.40x10-3 mol/L*s
3.21x10-3 mol/L*s
[O2]2
[O2]1
m
=1.10x10-2 mol/L
2.20x10-2 mol/Lm
; 2 = 2m m = 1
Do a similar calculation for the other reactant(s).
16-17
Sample Problem 16.3
PLAN:
SOLUTION:
Determining Reaction Orders 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.
16-18
Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data
0.080
0.0050
rate 2
rate 1
[NO2] 2
[NO2] 1
m
=k [NO2]
m2[CO]n2
k [NO2]m
1 [CO]n1
=
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
16-19
Table 16.3 Units of the Rate Constant k for Several Overall
Reaction Orders
Overall Reaction Order Units of k (t in seconds)
0 mol/L*s (or mol L-1 s-1)
1 1/s (or s-1)
2 L/mol*s (or L mol -1 s-1)
3 L2 / mol2 *s (or L2 mol-2 s-1)
16-20
Integrated Rate Laws
rate = -[A]
t= k [A]
rate = -[A]
t= k [A]0
rate = -[A]
t= k [A]2
first order rate equation
second order rate equation
zero order rate equation
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
16-21
Sample Problem 16.5
PLAN:
SOLUTION:
Determining the Reactant Concentration at a
Given Time
PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order
reaction, with the very high rate constant of 87 s-1, to two
molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00 M, 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]= (87 s-1)(0.010 s)
[C4H8] = 0.83 mol/L
ln[C4H8]0
[C4H8]t= kt
(a)
(b) [C4H8]0 - [C4H8]t
[C4H8]0
=2.00 M - 0.87 M
2.00 M= 0.58
16-22
Figure 16.7 Integrated rate laws and reaction orders.
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
[A]t = -kt + [A]0
16-23
Figure 16.8 Graphical determination of the reaction order for the
decomposition of N2O5.
16-24
Figure 16.9 A plot of [N2O5] vs. time for three half-lives.
t1/2 =
for a first-order process
ln 2
k
0.693
k=
16-25
Sample Problem 16.6
PLAN:
SOLUTION:
Using Molecular Scenes to Determine Half-Life
PROBLEM: The 1st order process of compound A (red) converting to
compound B (black) is depicted at 0.0 s and 30.0 s:
(a) Find the half-life, t1/2, of the reaction. (b) Calculate the rate constant, k.
(c) Draw a scene that represents the reaction mixture at 2.00 min.
Number of spheres represents concentration. Half-life is constant for
a 1st order reaction and the elapsed time when half the red spheres
turn black.
(a) At t = 0, 8 A and 0 B. At t = 30.0 s, 6 A and 2 B. For 4 A and
4 B, t = 60.0 s = t1/2.
(b) t1/2 = 0.693/k; k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10-2 s-1
(c) The 2.00 min is 120. s or two half-lives represented by the
above scene.
(c)
16-26
Sample Problem 16.7
PLAN:
SOLUTION:
Determining the Half-Life of a First-Order Reaction
PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its
60o bond angles allow poor orbital overlap, its bonds are weak.
As a result, it is thermally unstable and rearranges to propene at
1000oC via the following first-order reaction:
CH2
H2C CH2(g)
H3C CH CH2 (g)
The rate constant is 9.2 s-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, t1/2 = 0.693
k, to find the half-life.
One-quarter of the initial value means two half-lives have passed.
t1/2 = 0.693/9.2 s-1 = 0.075 s(a) 2 t1/2 = 2(0.075 s) = 0.150 s(b)
16-27
Table 16.4 An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero Order First Order Second Order
Plot for straight line
Slope, y intercept
Half-life
Rate law rate = k rate = k[A] rate = k[A]2
Units for k mol/L*s 1/s 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 = t
k, [A]0 -k, ln [A]0k, 1/[A]0
[A]0/2k ln 2/k 1/k[A]0
16-28
Figure 16.10 Dependence of the rate constant on temperature.
16-29
The Arrhenius Equation
The Effect of Temperature on Reaction Rate
RTEek
/aA
lnk = lnA - Ea/RT
lnk2
k1
=Ea
R-
1
T2
1
T1
-
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
T is the Kelvin temperature
A is the collision frequency factor
16-30
Figure 16.11 Graphical determination of the activation energy.
ln k = (-Ea/R )(1/T) + lnA
16-31
Sample Problem 16.8
PLAN:
SOLUTION:
Determining the Energy of Activation
PROBLEM: The decomposition of hydrogen iodide,
2HI(g) H2(g) + I2(g)
has rate constants of 9.51x10-9 L/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.
lnk2
k1
=Ea
-R
1
T2
1
T1
- Ea = - R lnk2
k1
1
T2
1
T1
-
-1
1
600 K
1
500 K-ln
1.10x10-5 L/mol*s
9..51x10-9 L/mol*s
Ea = - (8.314 J/mol*K)
Ea = 1.76x105 J/mol = 176 kJ/mol
16-32
Figure 16.12
Information sequence to determine the kinetic parameters of a reaction.
Series of plots
of concentra-
tion vs. time Initial
rates 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 at
varied T
Determine slope
of tangent at t0 for
each plot
Compare initial
rates when [A]
changes and [B] is
held constant and
vice versa
Substitute initial rates,
orders, and concentrations
into 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 at
varied T
16-33
Figure 16.13 The dependence of number 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
Why concentrations Are
Multiplied in the Rate Law ?
16-34
Figure 16.14
The effect of temperature on the distribution of collision
energies.Energy required to
activate the molecules
into a state from which
reactant bonds can
change into product
bonds
The temperature rise
enlarges the fraction of
collisions with enough
energy to exceed the Ea
16-35
Table 16.5 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)
25oC (298 K) 1.70x10-9
35oC (308 K) 3.29x10-9
45oC (318 K) 6.12x10-9
Increasing Ea
by 25 kJ/mol Decreased
16-36
Figure 16.15 Energy-level diagram for a reaction.
REACTANTS
PRODUCTS
ACTIVATED STATE
Colli
sio
n E
ne
rgy
Colli
sio
n E
ne
rgy
Ea (forward)
Ea (reverse)
The forward reaction is exothermic because the
reactants have more energy than the products.
16-37
Figure 16.16 An energy-level diagram of the fraction of collisions
exceeding Ea.
LargerSmaller
16-38
• In both reaction direction, a larger fraction of collision
exceeds the activation energy at the higher temperature, T2;
higher T increase reaction rate
16-39
Figure 16.17
The importance of molecular orientation to an effective collision.
NO + NO3 2 NO2
A is the frequency factor
A = pZ where Z is the collision frequency
p is the orientation probability factor
16-40
Figure 16.18
Nature of the transition state in the reaction between CH3Br and OH-.
CH3Br + OH- CH3OH + Br -
transition state or activated complex
16-41
Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.
Transition state/activated complex
Extremely unstable species
(has very high potential energy)
Forms only if the molecules collide
In an effective orientation and with
energy than Ea
Thus, the Ea is the quantity needed
To stretch and deform bonds in
order to reach the transition state
16-42Reaction progress
Po
ten
tia
l E
ne
rgy
Sample Problem 16.9
SOLUTION:
Drawing Reaction Energy Diagrams and
Transition States
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 the
products and the transition state is slightly higher than the
reactants.
O3+O
2O2
Ea= 19 kJ
Hrxn = -392 kJ
Ea(rev)= (392 + 19) kJ =
411kJ
OO
OO
breakingbond
formingbond
transition state
16-43
Table 16.6 Rate Laws for General Elementary Steps
Elementary Step Molecularity Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = [A]
Rate = k[A]2
Rate = k[A][B]
Rate = k[A]2[B]
REACTION MECHANISMS
16-44
Sample Problem 16.10
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)
16-45
The Rate-Determining Step of a Reaction Mechanism
The overall rate of a reaction is related to the rate of the slowest, or
rate-determining step.
Correlating the Mechanism with the Rate Law
The elementary steps must add up to the overall equation.
The elementary steps must be physically reasonable.
The mechanism must correlate with the rate law.
16-46
Figure 16.21
Reaction energy diagram for the two-step reaction of NO2 and F2.
2NO2(g) + F2(g) 2NO2F(g)
Accepted mechanism
NO2(g) + F2(g) NO2F(g)
+ F(g) [slow]
NO2(g) + F(g) NO2F(g)
NO2(g) + NO2(g) + F2(g) +
F(g) NO2F(g) + NO2F(g)
+ F(g)
16-47
CATALYSTS
•Each catalyst has its own specific way of functioning.
•In general a catalyst lowers the energy of activation.
•Lowering the Ea increases the rate constant, k, and
thereby increases the rate of the reaction.
•A catalyst increases the rate of the forward AND the
reverse reactions.
•A catalyzed reaction yields the products more quickly,
but does not yield more product than the uncatalyzed
reaction.
•A catalyst lowers Ea by providing a different mechanism,
for the reaction through a new, lower energy pathway.
16-48
Figure 16.22
Reaction energy diagram of a catalyzed and an uncatalyzed process.
16-49
Figure 16.23
Mechanism for the catalyzed hydrolysis of an organic ester.
H++ R C
O
O
R'
R C
O
O
R'
Hfast R C
O
O
R'
H
R C
O
O
R'
H
R C
O
O
R'
H
resonance hybrid
resonance forms
R C
O
O
R'
H
H
O H R C
O
O
R'
H
H
O H
slow, rate-determining
step
R C
O
OH
R'
H+
O Hall fast
16-50
16-51
Figure 16.24 The metal-catalyzed hydrogenation of ethene.
H2C CH2 (g) + H2 (g) H3C CH3 (g)
16-52
Figure 16.24 The metal-catalyzed hydrogenation of ethene.
16-53
Tools of the Laboratory
Figure B16.1 Spectrometric monitoring of a reaction.
16-54
Figure B16.2
Figure B16.3
Conductometric monitoring
of a reaction.
Manometric monitoring of a
reaction.
Tools of the Laboratory
16-55
Figure B16.4
The widely separated amino acid groups that form an active site.
16-56
Figure B16.5 Two models of enzyme action.
16-57
Figure B16.6 The increasing size of the Antarctic ozone hole.
O2 2O
O + O3 2O2 (ozone breakdown)
O + O2 O3 (ozone formation)