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Chapter 14-Chemical Kinetics
Tour of the Ozone Hole
http://www.atm.ch.cam.ac.uk/tour/
16 September INTERNATIONAL DAY FOR THE PRESERVATION
OF THE OZONE LAYER
The Ozone Hole of 2008 is larger than in 2007
Geneva, 16 September 2008 (WMO) - “After decades of
chemical attack, it may take another 50 years or so for the
ozone layer to recover fully. As the Montreal Protocol has
taught us, when we degrade our environment too far, nursing
it back to health tends to be a long journey, not a quick fix”,
said Ban Ki-moon, Secretary-General of the United Nations,
on the occasion of the International Day for the Preservation
of Ozone Layer today. According to the World Meteorological
Organization (WMO), the 2008 Antarctic ozone hole will be
larger than the one of 2007. The observed changes in the
stratosphere could delay the expected recovery of the ozone
layer. It is therefore vital that all Member States with
stratospheric measurement programmes continue to support
and enhance these measurements.
O3 concentration is
reported in Dobson
Units (DU)
1DU =2.69E15 O3
molecules/cm2
In September, 2006 the
largest ozone hole ever
was observed.
Temperature
inversions and wind
direction changes
dramatically
influence air quality
in cities like Salt
Lake…
Photo-chemical Smog
NO(g) + O3(g) → NO2(g) + O2(g)
In 1998, a total of
24.5 megatons of
nitrogen oxide (NOx)
compounds were
released to the
atmosphere.
NOx compound are
produced primarily in
the combustion of
hydrocarbons
N2(g) + O2(g) → 2 NO(g)
DH° = + 180.6 kJ
2 NO(g) + O2(g) → 2 NO2(g)
DH° = - 114.2 kJ
NO2(g) + hn → NO(g) + O(g)
photochemical
O2(g) + O(g) → O3(g)
Ozone production
Chemical Kinetics looks at both the chemical
mechanism as well as the rate at which reactions
occur.
MECHANISM - how reactants combine to form products.
This includes the order that bonds are broken and formed
and the possible production of any intermediates.
REACTION RATE - a measure of how fast a reaction occurs.
This is measured by monitoring the decrease in
concentration of the reactants and/or the increase in the
concentration of the products. Conditions of temperature,
concentration or pressure, the presence of catalyst, etc.
that affect the reaction rate may also be specified.
Chemical Reactions usually involve molecular
collisions that result in bonds being broken or made.
Both of these reactions are bimolecular…i.e. two
molecules collide in order for the reaction to occur.
CONDITIONS FOR BIMOLECULAR AND HIGHER
REACTIONS:
1. Two Or Molecules Must Collide.
2. The Colliding Molecules Must Be In The Proper Orientation.
3. The Colliding Molecules Must Have Enough Energy to React.
The Lewis structure of NO2 indicates that the nitrogen
atom has an unpaired electron. Two NO2 molecules
combine by using their unpaired electrons to form an
N-N bond.
Chemical Reaction Mechanism
In the reverse of the previous reaction, a unimolecular
elementary reaction may involve bond breakage. If an
N2O4 molecule possesses enough energy (from heat or
light), molecular vibrations can break the N-N bond to
produce two NO2 molecules.
When 3 molecules collide to form chemical product the
reaction mechanism would be called ter-molecular.
The reaction between O3 and NO is believed to
occur by a mechanism that consists of the single
bimolecular step illustrated here in a molecular
view.
A chemical reaction rate is the change in concentration
of a reactant or product during a particular time interval
for the reaction
N2(g) + O2(g) → 2 NO(g)
N2(g) + O2(g) → 2 NO(g)
Reaction Rate Tutorial
»PC version
Learn to calculate the average and instantaneous rate
from the rate expression and concentration vs. time data.
Includes practice exercises.
•
(a) To calculate the average rate of reaction,
determine how many moles are consumed during the
time interval and divide by the time:
• Dn = (0.25 mol/L) – (0.50 mol/L) = – 0.25 mol/L;
•
Rate = M/s = 0.25 M/30. s = 8.33E-3
• Round to two significant figures: Rate = 8.3E-3 M/s
• NOTE UNITS of REACTION RATES
• (b) The rate for any particular reagent is the
coefficient for that reagent times the rate of reaction:
Rate(NH3) = (2/3)(8.3-3 M/s) = 5.6 x 10-3 M/s
• (c) The concentration of any particular reagent is its
initial concentration minus the change during the time
interval:
• Change = (Coeff)(Rate)(time) =
–(1/3)(8.3E-3M/s)(30 s) = –8.3 x 10-2 M
•
Concentration = (1.25 M) – (0.083 M) = 1.17 M
Problem
An engineer is studying the rate of the Haber
synthesis:
N2(g) + 3 H2(g) → 2 NH3(g)
Starting with a closed reactor containing 1.25
mol/L of N2 and 0.5 mol/L of H2, the engineer
find that the H2 concentration has fallen to
0.25 M after 30 seconds.
(a) What is the average rate of reaction over
this time.
(b) What is the ave. rate of production of
NH3.
(c) What is the N2 concentration after 30s?
(a) What is the average rate of reaction over this time.
(b) What is the ave. rate of production of NH3.
(c) What is the N2 concentration after 30s?
2 NO2 → 2 NO + O2
A plot of [NO2], [O2], and [NO] as a function of time
(seconds) for the decomposition reaction of NO2. The
concentration data are shown in the table.
Reaction Rate for NO2 Decomposition
0.0012
Conc (mol/L)
0.001
0.0008
Initial Rate
0.0006
Instantaneous Rate
0.0004
Average Rate
0.0002
0
0
50
100
150
Time (s)
200
250
300
Instantaneous Rates
The rate at t=0 is the instantaneous
initial rate. This is the most
common way of reporting rates in
the laboratory
RATE TYPES
1. INSTANTANEOUS RATE - the rate at a particular point in
time during the reaction. The instantaneous rate is the slope
of a tangent at any point in time along the curve.
2. INITIAL RATE - the instantaneous rate at t = 0. This is often
the rate used in a discussion of the kinetics of a reaction.
3. AVERAGE RATE - the rate averaged over a particular time
interval. The average rate determined by D[conc]/Dt under
estimates the true average rate,
(D[conc]/n)/ Dt, because the rate varies over time.
Instantaneous Rates
Tropospheric ozone is rapidly consumed in
many reactions, including the following.
O3 + NO → NO2 + O2
Use the following data to calculate the
instantaneous rate of the preceding
reaction at t = 0.000 s and t = 0.102 s.
Time (s)
[NO](M)
0.000
2.0E-8
0.011
1.8E-8
0.027
1.6E-8
0.052
1.4E-8
0.102
1.2E-8
FACTORS THAT AFFECT REACTION RATES
1. Concentration Of The Reacting Species. For reversible
reactions, this includes reactants and products.
2. Temperature. Increasing the temperature increases the rate of
endothermic and exothermic reactions. Other forms of energy,
such as light, electrical, etc. , may also increase a reaction rate.
3. Catalyst. A catalyst increases reaction rates by changing the
mechanism and lowering the reaction energy barrier.
4. Surface Area Of Reactants. The greater the contact surface
area, the faster the reaction rate.
The rate of a chemical reaction increases with
increasing concentration because the reactants
(NO + O2) are more likely to collide.
It follows:
reaction rate ~ [reactants]
Or that
reaction rate = k[reactants]n
Note that the units of “k” (the rate constant) are
different depending on n, the “order” of the
reaction
14.48. Compounds A and B react to give a single product, C.
Write the rate law for each of the following cases and determine
the units of the rate constant by using the units M for
concentration and s for time:
a.
The reaction is first order in A and second order in B.
b.
The reaction is first order in A and second order overall.
c.
The reaction is independent of the concentration of A and
second order overall.
d.
The reaction is second order in both A and B.
14.53.Rate Laws for Destruction of Tropospheric Ozone
The reaction of NO2 with ozone produces NO3 in a secondorder reaction overall:
NO2(g) + O3(g) → NO3(g) + O2(g)
a.
b.
Write the rate law for the reaction if the reaction is first
order in each reactant.
The rate constant for the reaction is 1.93  104 M–1s–1
at 298 K. What is the rate of the reaction when
[NO2] = 1.8  10–8 M and [O3] = 1.4  10–7 M?
c.
d.
What is the rate of the appearance of NO3 under these
conditions?
What happens to the rate of the reaction if the
concentration of O3(g) is doubled?
b.
The rate constant for the reaction is 1.93  104 M–1s–1 at
298 K. What is the rate of the reaction when
[NO2] = 1.8  10–8 M and [O3] = 1.4  10–7 M?
c.
d.
What is the rate of the appearance of NO3 under these
conditions?
What happens to the rate of the reaction if the
concentration of O3(g) is doubled?
Table 14.3 Effect of Reactant Concentrations on Initial
Rates of: 2 NO(g) + O2(g) → 2 NO2(g)
Expr. # [NO]0
[O2]0
Initial Rxn Rate, NO
(M/s)
1
0.0100
0.0100
2.0 E-6
2
0.0100
0.0050
1.0 E-6
3
0.0050
0.0100
5.0 E-7
4
0.0050
0.0050
2.5 E-7
What is the rate equation (rate law) for this
reaction?
The single experiment approach
For reactions such as: O3(g) → O2(g) + O(g)
The rate is found to be first order in ozone
rate = k[O3]
The mathematical solution can be found by
integrating this expression.
What are the units for “k”?
The First Order integrated rate law is:
ln([O3]/[O3]0) = - kt
Problem
The decomposition of N2O5 (g) is 1st Order.
(a) Write the rate equation.
(b) If 2.56 mg of N2O5 is present initially,
and 2.50 mg remain after 4.26 min at 55
C. Calculate the rate constant, k, for this
process.
N2O(g) → N2(g) + 1/2 O2(g)
Rate = k[N2O]
t½ = 1 s
14C
t½
14_10.jpg
→ 14N + 0-1e
= 5730 yr
Half-life Problem
The half-life for a first order reaction at
550°C is 85 seconds. How long would it
take for 23% of the reactant to
decompose?
At high temperatures, the
reaction:
2 NO2(g) → 2NO(g) + O2(g)
is second-order in NO2, i.e.
Rate = k[NO2]2
The integrated rate law
for a 2nd order reaction
is:
[NO2]-1t = kt + [NO2]-10
What are the units for “k”?
14.64. Two structural isomers of ClO2 are shown:
The isomer with the Cl–O–O skeletal arrangement is unstable
and rapidly decomposes according to the reaction
2ClOO(g) → Cl2(g) + 2O2(g).
The following data were collected for the decomposition of ClOO
at 298 K:
Time (μs)
[ClOO] (M)
0.00
1.76  10–6
0.67
2.36  10–7
1.3
3.56  10–8
2.1
3.23  10–9
2.8
3.96  10–10
Determine the rate law for the reaction and the value of the
rate constant at 298 K. What is the half-life for the reaction?
Is the reaction 1st or 2nd order in Cl-O-O?
Plot of Concentration (M) versus Time
2.00E-06
1.80E-06
1.60E-06
1.40E-06
[COO]
1.20E-06
1.00E-06
Conc. mol/L
Power (Conc. mol/L)
8.00E-07
6.00E-07
4.00E-07
2.00E-07
0.00E+00
0
0.5
1
1.5
Time (us)
2
2.5
3
Is the reaction 1st or 2nd order in Cl-O-O? Second-Order Plot…
1/[COO]
3.000E+09
2.500E+09
2.000E+09
[COO]-1
1.500E+09
1/[COO]
1.000E+09
5.000E+08
0.000E+00
0
0.5
1
1.5
Time (us)
2
2.5
3
Is the reaction 1st or 2nd order in Cl-O-O? First-Order Plot…
0
0
0.5
1
1.5
2
2.5
3
-5
-10
y = -3x - 13.25
ln[COO]
Series1
Linear (Series1)
-15
-20
-25
Time (us)
Problem Chapter 14
The condensation reaction of butadiene has a rate
constant of 0.93 L/mol۰min. If the initial
concentration of C4H6 is 0.240 M, find:
(a) the time at which the concentration will be 0.100 M
(b) the concentration after 3.5 hr.
•(a) The units of the rate constant indicate that this is a
second-order reaction, so Rate = k[C4H6]2 and
•
1/[A] – 1/[A]o = kt ;
•The problem states that k = 0.93 M-1 min-1.
•
(a) kt = (10.0 – 4.17) M-1 = 5.83 M-1;
•
t = 6.3 min;
•(b) 1/[A] = 1/0.24M + (0.93 M-1 min-1)(25 min)
= (4.17 + 23.25) M-1
•
= 27.42 M-1
•
[A] = 3.6E-2 M.
Reaction Order Tutorial
»PC version
Use interactive graphs to explore how reaction rate varies
with concentration of reactant. Example problems outline
the steps for determining the rate law and rate constant
from concentration and initial rate data. Includes practice
problems.
58. Hydroperoxyl radicals (HO2) react rapidly with
ozone in this elementary reaction
HO2 + O3 → OH + 2 O2
Determine the pseudo-first-order rate constant and the
second-order rate constant for this reaction from the data
Time(ms)
[HO2]
[O3]
0
3.2E-6
1.0E-3
10
2.9E-6
1.0E-3
20
2.6E-6
1.0E-3
30
2.4E-6
1.0E-3
80
1.4E-6
1.0E-3
What is the
molecularity of
the reaction?
First order Plot
y = -0.0103x - 12.647
R2 = 0.9991
-12.6
0
20
40
60
80
100
-12.7
-12.8
Slope = -k’ (pseudo 1storder)
-12.9
Ln[HO2]
-13
-13.1
Linear
-13.2
-13.3
-13.4
-13.5
-13.6
time(ms)
Problem Solution
From the plot, k’ = 1.03E-2 ms-1
And the pseudo first-order rate equation is:
Rate = k’[HO2]
The overall rate must be”
rate = k[O3][HO2]
So that
k’ = k[O3]
And
k = k’/[O3] = 1.03E-2 ms-1/1.0E-3M
= 1.0E+1 ms-1M-1
Reaction Mechanism
Overall reaction:
2 NO2(g) → 2 NO(g) + O2(g)
Reaction mechanisms…elementary reactions
Step 1)
NO2 + NO2 → NO + NO3
Step 2)
NO3 → NO + O2
Overall)
2 NO2 → 2 NO + O2
NO3 is a reaction intermediate as it does not
appear in the overall reaction
Reaction mechanisms…rate determining steps
Step 1)
NO2 + NO2 → NO + NO3
(slow)
Step 2)
NO3 → NO + O2
(fast)
Overall)
2 NO2 → 2 NO + O2
Rate (1) = k1[NO2]2
Rate (2) = k2[NO3]
The slowest step in the mechanism is usually
Rate Determining.
For this example, the mechanism predicts that
the experimentally observed rate will be second
order in NO2.
LINKING MECHANISMS AND RATE LAWS
1. The mechanism is one or more elementary reactions that
describes how the chemical reaction occurs.
These elementary reactions may be unimolecular,
bimolecular, or (very rarely) termolecular.
2. The sum of the individual steps in the mechanism must
give the overall balanced chemical equation.
3. The reaction mechanism must be consistent with the
experimental rate law.
If the rate law predicted by the proposed mechanism differs
from the experimental rate law, the proposed mechanism is
wrong.
If the rate law predicted by the proposed mechanism matches
the experimental rate law, the proposed mechanism is a
possible description of how the reaction proceeds, but must
be verified by experiments.
Reaction Mechanisms Tutorial
»PC version
Learn to calculate the rate expression of a multi-step
reaction from its elementary steps by identifying the
rate-determining step. Includes practice exercises.
Problem 70. The rate laws for the thermal and
photochemical decomposition of NO2 are different.
Which of the following mechanisms can be
attributed to thermal versus photochemical
decomposition. Given:
Rate (thermal) = k[NO2]2
Rate (photo) = k[NO2]
a.
NO2 + NO2 → N2O4
slow
N2O4 → N2O3 + O
fast
N2O3 + O → N2O2 + O2
fast
N2O2 → 2 NO
fast
Problem 103. The rate laws for the thermal and
photochemical decomposition of NO2 are different. Which of
the following mechanisms can be attributed to thermal
versus photochemical decomposition. Given:
Rate (thermal) = k[NO2]2
Rate (photo) = k[NO2]
a.
b.
c.
Step 1:
Step 2:
NO2(g) → NO(g) + O(g)
O(g) + NO2(g) + O2(g)
slow
fast
Problem 69b
step1
NO2 + NO2 → N2O4
fast
step2
N2O4 → NO + NO3
slow
step3
NO3
fast
→ NO + O2
For this case, the second step is Rate Determining, so that
the overall rate would be:
rate = k2[N2O4]
However, since N2O4 is a reaction intermediate and not
generally observed, we must substitute NO2 for it. We can
do this by assuming that step1 is both fast and reversible.
Then the rate forward for step1 is:
rate forward = k1[NO2]2
And the rate of the reverse reaction is:
rate reverse = k-1[N2O4]
Note that the forward and reverse reactions have different
rate constants.
step1
NO2 + NO2 → N2O4
fast
step2
N2O4 → NO + NO3
slow
step3
NO3
fast
→ NO + O2
If it is then assumed that both the forward and back
reactions quickly reach equilibrium, i.e. we set the rate
forward equal to the rate reverse,
rate forward= k1[NO2]2 = k-1[N2O4] = rate reverse
Then we can algebraically solve for the concentration of the
intermediate…N2O4
[N2O4] = k1/k-1[NO2]2
And substitute this value into the rate-determining step2
rate = k2[N2O4]
= k2(k1/k-1)[NO2]2
= kobserved[NO2]2
Now the mechanism and rate law are consistent with
Reaction Rates and
Temperature
The rate constant, k,
depend on temperature
as:
k = Ae-(Ea/RT)
Where the parameter
“A” is a frequency
factor, and Ea is the
activation energy for
the reaction.
The frequency factor is the
number of collisions per sec
times the probability that a
collision has an effective
orientation
The Arrhenius Equation
k = Ae-(Ea/RT)
Was developed from the
observed relationship between
rate (k) and temperature.
The theory postulates an
Activation Energy (Ea) which is
the energy barrier that must be
overcome before two molecules
can react.
Svante August Arrhenius
Activation energy
R*
R
Activated
complex
Reaction Rates and Temperature. Effect of molecular
orientation:
O3 + NO → NO2 + O2
For reactions that
proceed via more
than one step,
there may be
more than one Ea
linked to forming
the activated
complex
Reaction Mechanisms
2 NO2(g) → 2 NO(g) + O2(g)
The elementary reaction step
with the larger activation energy
will be rate determining.
NO + O3 → NO2 + O2 DH < 0
For the reverse reaction:
NO2 + O2 → NO + O3 DH > 0
Which process has the larger rate constant, k?
Exothermic
Endothermic
To determine Ea, the rate of reaction
must be measured over several
temperatures.
Ln(k) = -Ea/(RT) + lnA
Ammonium cyanate (NH4NCO) decomposes to
urea(NH2CONH2). The following data were obtained at
50°C.
Time(hr)
0
1.0
Conc.(M) 0.500 0.375
2.0
3.0
5.0
7.0
9.0
0.300
0.250
0.188
0.150
0.125
At 25°C the concentration falls from 0.500 M to 0.300
M in 6.0 hrs.
A) Determine the rate Law
B) Determine the rate constant at 50°C.
C) Determine the activation energy.
Problem
Values of the rate constant of the reaction:
N2 + O2 → 2NO
Are as follows:
A) Calculate Ea
B) Calculate A (freq. factor)
c) Calculate k at T=300K
T(K)
2000
2100
2200
2300
2400
k (M-½ s-1)
318
782
1770
3733
7396
Problem (in Excel)
12
11
10
9
8
ln(k)
7
6
5
4
y = -37758x + 24.641
R2 = 1
3
2
1
0
0.0004
0.00042
0.00044
0.00046
1/T(K)
0.00048
0.0005
0.00052
Arrhenius Equation Tutorial
»PC version
This tutorial explores how energy, rate constant, and the
effects of temperature and orientation are related.
Includes practice exercises.
Collision Theory Tutorial
»PC version
Explore the effects of temperature, orientation of
reactant molecules, and catalysts on reaction rates.
Includes practice exercises.
Rate of reaction versus reaction “spontaneity”.
The rate of an endergonic reaction (DG<0) may be
greater than that for an exergonic reaction (DG<0)
Catalysts increase the rate of reaction by lowering
the activation energy. Note: a catalyst will not
change DE (or DG) for the reaction.
2O3(g) ↔ 3O2(g)
The activation energy diagram for the reaction between
O3 and O. Note the Cl atoms remain.
Catalytic converters reduce emissions of NO (and CO)
by lowering the activation energy for decomposition
to N2 and O2. That is adsorption of NO on the Pt/Pl
surface weakens the N-O bond.
Catalytic Converter
EPA proposed
reduction in NOx
emissions by the
year 2007.
Table 14.7: Federal (EPA) Emissions Standards for
Automobile Exhaust
NOx(g/mi) CO(g/mi) Hydrocarbon(g/mi)
Pre-1976
3.5-7.0
83-90
13-16
1973
3.1
15.0
1.5
1991
1.0
3.4
0.41
2004
0.2
1.7
0.125
W. W. Norton & Company
Independent and Employee-Owned
This concludes the Norton Media Library
slide set for chapter 14
Chemistry
The Science in Context
by
Thomas Gilbert,
Rein V. Kirss, &
Geoffrey Davies

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