Chapter Outline - Western Oregon Universitypostonp/ch223/pdf/Ch13-s15.pdf · 2019-04-08 ·...
Transcript of Chapter Outline - Western Oregon Universitypostonp/ch223/pdf/Ch13-s15.pdf · 2019-04-08 ·...
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• 13.1 Cars, Trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Cars, Trucks, and Air Quality
• Photochemical Smog: Mixture of gases formed when sunlight interacts with compounds produced in internal combustion engines
• Depends on chemical kinetics
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Important Reactions in Smog
1. N2(g) + O2(g) → 2 NO(g) ΔH= 180.6 kJ
2. 2 NO(g) + O2(g) → 2 NO2(g) ΔH= -114.2 kJ
3. NO2(g) NO(g) + O(g)
4. O2(g) + O(g) → O3(g)
5. O(g) + H2O(g) → 2 OH(g)
sunlight⎯⎯⎯→
Note: Products of some reactions are reactants in
other reactions.
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Variations of Smog Components
Photodecomposition
of NO2 leads to high
levels of O3 in the
afternoon.
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Why Study Kinetics?
• We can control the reaction conditions to
obtain product as quickly and economically
as possible
• To understand reaction mechanisms – a
study of kinetics sheds light on how a
reaction proceeds
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Factors that control reaction kinetics
• Chemical makeup of reactants and products
• Concentration of reactants
• Temperature
• Catalysts
• Surface area
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• 13.1 Cars, Trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Rates
If driving a car, rate = change in position
change in time= slope=
∆dist∆t =
d2−d1t2−t1
Time, hours →
Dis
tan
ce
, m
iles → (t2, d2)
(t1, d1)
rise = 𝑑2 − 𝑑1
run = 𝑡2 − 𝑡1
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Reaction Rates
rate = change in conc.change in time
= slope = ∆[A]∆t =
[A]2−[A]1t2−t1
Time (s, min, hr) →
Con
ce
ntr
atio
n, M
→
(t2, [A]2)
(t1, [A]1)
rise =
[𝐴]2−[𝐴]1
run = 𝑡2 − 𝑡1
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Reaction Rates:
rate of loss of reactant = negative slope
rate = change in conc.change in time
= slope = ∆[A]∆t =
[A]2−[A]1t2−t1
Time (s, min, hr) →
Con
ce
ntr
atio
n, M
→
(t2, [A]2)
(t1, [A]1)
rise =
[𝐴]2−[𝐴]1
run = 𝑡2 − 𝑡1
Here the slope is negative, so the
rate is multiplied by -1 to make
the rate positive (convention)
Rate = -∆[A]∆t
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For the reaction A → B
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8 9 10
Co
nc
en
tra
tio
n, M
Time, min
[A]t
[B]t
Note how the reaction
rate decreases over time
Here the slope is
negative
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Reaction Rates and
Stoichiometry
2A B
As an example, two moles of A disappear for each
mole of B that is formed, so the rate of disappearance
of A is twice the rate of appearance of B.
Mathematically: −∆[𝐴]
∆𝑡= 2
∆[𝐵]
∆𝑡or
−1
2
∆[𝐴]
∆𝑡=
∆[𝐵]
∆𝑡
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Reaction Rates and
Stoichiometry (cont’d)
2A 3BA trickier example: two moles of A
disappear and three moles of B are
formed. First divide both sides by 3 --
Mathematically: −∆[A]∆t
= 23
∆[B]∆t
or
23
A B So now the rate of disappearance of A is
2/3 times the rate of appearance of B.
−12
∆[A]∆t
= 13
∆[B]∆t
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aA + bB cC + dD
In general -
Reactants have
the minus signproducts
= 1
𝑐
∆[𝐶]
∆𝑡=
1
𝑑
∆[𝐷]
∆𝑡= -
1
𝑏
∆[𝐵]
∆𝑡-1
𝑎
∆[𝐴]
∆𝑡
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Actual Reaction Rates
N2(g) + O2(g) → 2 NO(g)Smog
reaction #1
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Average Rates
N2(g) + O2(g) → 2 NO(g)
Average Rate: Change in concentration of
reactant or product over a specific time interval
average rate =∆[NO]
∆t
= ([NO]f − [NO]i)
(tf − ti)
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Practice: Reaction Rates
• Calculate the average reaction rate for the
formation of NO between 5 s and 10 s, and
then between 25 s and 30 s, using the data
below: N2(g) + O2(g) → 2 NO(g)
(a)
(b)
Δ NOAverage Rate =
Δt
(a)
(b)
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Instantaneous Reaction Rates
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Instantaneous Reaction Rates
• Instantaneous Rate:
• Reaction rate at a particular instant
• Determined graphically as tangential slope
of concentration vs. time plot
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• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Effect of Concentration on
Reaction Rates
A typical plot of reactant concentration versus time. The
rate depends on the number of molecular collisions
which depends on the concentration of reactants.
a) Initial rate (will be used in the “Method of Initial
Rates”)
b) midpoint
c) end
The rate approaches zero
as the number of reactant
molecules decreases =
concentration effect
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Reaction Order and Rate Constants
We observe in chemical reactions that rates increase as
the reactant concentrations increase. The rate is found to
be proportional to the reactant concentrations raised to
some experimentally determined power -
General Rate Law: Rate = k [A]m [B]n
o k = rate constant
o m and n are the reaction orders with respect to reactants A and B
o (m + n ) = overall order
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Overall Order
The sum of the exponents for each reactant in the Rate Law.
rate = k[A] 1st order in A, 1st order overall
rate = k[A][B] 1st order in A and B, 2nd order overall
rate = k[A]2[B] 2nd order in A, 1st order in B, 3rd order overall
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The Method of Initial Rates
An experimental procedure for determining the
overall reaction order by systematically varying
the concentrations of each reactant
Rate = k [A]m [B]n
Exp. # [A], M [B], M Rate,
Ms-1
1. 0.10 0.10 1 x 10-5
2. 0.10 0.20 2 x 10-5
3. 0.20 0.20 8 x 10-5
Example experimental setup:
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2 NO(g) + O2(g) → 2 NO2(g)
rate = k [NO]m [O2]n
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Systematically varying the concentrations
of each reactant, e.g., the exponent n -
Hold the concentration of one reactant while doubling,
tripling, etc the other concentration of the other reactant:
n: [NO] is held constant. [O2] doubles and the
rate doubles. Therefore n = 1
rate = k [NO]m [O2]n
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Systematically varying the concentrations
of each reactant, e.g., the exponent m -
rate = k [NO]m [O2]n
m: [O2] is held constant. [NO] doubles and
the rate quadruples. Therefore m = 2
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2 NO(g) + O2(g) → 2 NO2(g)
Solve for the rate constant k by using the data from any of
the experiments, e.g. experiment #1 -
rate = k [NO]2 [O2]
k =rate
[NO]2[O2]
k =1 x 10−6 M s−1
0.010 𝑀 2 [0.010 M]
k = 1.0 M-2 s-1
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Experimentally the rate = k [F2][ClO2]
Characteristics of Rate Laws
• Rate laws are always determined experimentally.
• Reaction order is always defined in terms of reactant
(not product) concentrations.
• The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation, e.g.
1
F2 (g) + 2 ClO2 (g) 2FClO2 (g)
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Equations for determining the exponents:
e.g. rate = k [A]m [B]n
Experiment [A], M [B], M rate
1. [A]1 [B]1 rate1
2. [A]1 [B]2 rate2
rate1rate2
=[B]1[B]2
n
Want to solve for n, so take the log of both sides first -
lograte1rate2
= log[B]1[B]2
n
rate1rate2
=[B]1[B]2
n
and since
log Xn = n log X
lograte1rate2
= n log[B]1[B]2
n=
lograte1rate2
log[𝐵]1[𝐵]2
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Sample Exercise 13.3 - Deriving a Rate
Law From Initial Reaction Rate Data
N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]
n
n
m
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N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]
n
n =
lograte1rate2
log[O2]1[O2]2
=log
707500
log0.0200.010
=0.15
0.30= 1/2
Calculating n, the exponent on O2:
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Calculating m, the exponent on N2:
m =
lograte2rate3
log[N2]2[N2]3
=log
500125
log0.0400.010
=0.60
0.60= 1
N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]
n
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rate = k [N2][O2]1/2
Solve for the rate constant k by using the data
from any of the experiments, e.g. experiment #1 -
k =707 M s−1
0.040 [0.020]1/2k = 1.25 x 105 M-1/2 s-1
k =rate
[N2][O2]1/2
N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]
n
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• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
o Integrated Rate Laws: First-Order Reactions
o Reaction half-lives
o Integrated Rate Laws: Second-Order Reactions
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Relation Between Reactant Concentration
and Time: Integrated Rate Laws
I. First Order Kinetics: rate = -∆[A]t
∆t= k1[A]
II. Second Order Kinetics: rate = -∆[A]t
∆t= k2[A]2
0.000
0.020
0.040
0.060
0.080
0.100
0.120
0 10 20 30 40 50
[A] t→
time →
2nd order
1st order
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[A]t = concentration of A at some time = t
[A]o = concentration of A at time t = 0 (initial concentration)
k1 = first order rate constant
(units = s-1, min-1, h-1, etc)
A (+ other reactants) → products
First Order Integrated Rate Equation
rate = -∆[A]t
∆t= k1[A] the solution to this equation is:
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Some mathematics: ln[A]
[A]o= −k1t
exp ln[A]t[A]o
= exp −k1t
[A]t[A]o
= exp −k1t
[A]t =
Ao exp −k1t
Solving for [A]t Solving for time, t
ln[A]
[A]o= −k1t
t = −1k1
ln[A]
[A]o
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1st Order Kinetics Example – HW # 69
The following reaction is known to be 1st order with
a rate constant k = 6.32 x 10-4 s-1. How much N2O5
remains in solution after 1 hr if [N2O5]o = 0.50 M?
2 N2O5 → 4 NO2 + O2
[N2O5]t = [N2O5]o exp −k1t
[N2O5]t = (0.50 M) exp[-(6.32 x 10-4 s-1)(3600 s)]
[N2O5]t = (0.50 M)(0.103)
[A]t = 0.052 M
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Linearizing the 1st Order Equation
Linearization of an equation is a common method
used in science where useful information can be
obtained from the slope and intercept (e.g. measuring
the rate constant).
ln[A]t[A]o
= −k1t And since lnab
= ln (a) − ln (b)
ln [A]t − ln [A]o = −k1t
ln [A]t = −k1t + ln [A]o
y = m x + b
m = -𝑘1
b = ln [A]o
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Example Linearization
(spreadsheet on the webpage)
0
0.002
0.004
0.006
0.008
0.01
0.012
0 20 40 60 80 100 120 140 160 180 200
[A]t
→
time, s
Example Data
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120 140 160 180 200
[A]t
→
time, s
1st Order Linearization
time, s 1st order data
0 0.01
10 0.007408182
20 0.005488116
30 0.004065697
40 0.003011942
50 0.002231302
60 0.001652989
70 0.001224564
80 0.00090718
90 0.000672055
100 0.000497871
110 0.000368832
120 0.000273237
130 0.000202419
140 0.000149956
150 0.00011109
160 8.22975E-05
170 6.09675E-05
180 4.51658E-05
190 3.34597E-05
200 2.47875E-05
ln([
A] t
[A] t
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Integrated Rate Law: 1st-Order
3 2 O + OhO ⎯⎯→
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y = m x + b
Plot data after
taking the log
Fit to a
straight line
m = -6.93 x 10-3 s-1
m = -k1 = -6.93 x 10-3 s-1
k1 = 6.93 x 10-3 s-1
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• Half-Life (t1/2):
• The time in the course of a chemical reaction
during which the concentration of a reactant
decreases by half
• From integrated rate law, when [A]t = ½ [A]o:
• Rearrange: 1/2
0.693t =
k
0
Aln = - 0.693 = -
At kt
Half-Life: 1st Order Reactions
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Half-Life: 1st-Order Reactions
t1/2 (1) = t1/2 (2) = t1/2 (3)
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Sample Exercise 13.5: Calculating the Half-Life
of a First-Order Reaction.
The rate constant for the decomposition of N2O5 at a particular
temperature is 7.8 x 10-3 s-1. What is the half-life of N2O5 at that
temperature?
t1/2 =0.693
k
t1/2 =0.693
7.8 x 10−3 s−1
t1/2 = 89 s
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Some example half lives for First Order
processes -
Process Half life, t1/2
Rate constant,
k (s-1)
Radioactive decay of 238U 4.51 X 109 yr 4.87 X 10-18
Radioactive decay of 14C 5.73 X 103 yr 1.21 X 10-4
C12H22O11(aq) + H2O →
C6H12O6(aq) + C6H12O6(aq)
(glucose) (fructose)
8.4 h 2.3 X 10-6
HC2H3O2 + H2O → H3O+ +
C2H3O2-
8.9 X 10-7 s 7.8 X 105
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Second Order Kinetics
A (+ other reactants) → products
[A]t = concentration of A at some time = t
[A]o = concentration of A at time t = 0 (initial concentration)
k2 = second order rate constant (units = M-1 s-1, M-1 min-1, etc)
rate = -∆[A]t∆t
= k2[A]2 the solution to this equation is:
1[A]t
= k2t +1
[A]o
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Solving for [A]t Solving for time, t
Some mathematics:1
[A]t= k2t +
1[A]o
k2t =[A]o− [A]t[A]t[A]o
t =1
k2
[A]o− [A]t[At][A]o
k2t =1
[A]t-
1[A]o
1[A]t
= k2t +1
[A]o
1
[A]t=
[A]ok2t + 1
[A]o
[A]t=[A]o
[A]ok2t+1
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Linearizing the 2nd Order Equation
y = m x + b
m = k2
b = 1
[A]o
1[A]t
= k2t +1
[A]o
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ln [NO2] plot: curved
1/[NO2] plot: straight line
slope = k2; intercept = 1/[NO2]o
How to graphically distinguish between 1st and 2nd order kinetics
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Sample Exercise 13.6: Distinguishing
between First and Second order reactions
Chlorine monoxide accumulates in the stratosphere above
Antarctica each winter and plays a key role in the formation of
the ozone hole above the South pole each spring. Eventually
ClO decomposes according to the reaction:
2 ClO(g) → Cl2(g) + O2(g)
The kinetics of this reaction were studied in a laboratory
experiment at 298 K, and the data are shown in Table
13.10(b). Determine the order of the reaction, the rate law,
and the value of the rate constant k.
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2 ClO(g) → Cl2(g) + O2(g)
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• because the plot of 1/[ClO]
is linear, the reaction is 2nd
order
• So the rate law = k2[ClO]2
• from the straight-line fit:
m = 7.24 x 106
y = 7.24E+06x + 6.63E+07
Results:
m = k2 = 7.24 x 106 M-1s-1
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Half Life for Second Order Reactions
When t = t½ ,
then [A]t1/2 = [A]o/2
1[A]t
= k2t +1
[A]o
k2t =1
[A]t-
1[A]o
k2t1/2 =1
[A]o/2-
1[A]o
k2t1/2 =2
[A]o-
1[A]o
k2t1/2 =1
[A]o
t1/2 =1
k2[A]o
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Sample Exercise 13.7: Calculating the
half-life of Second order reactions
Calculate the half-life of the second-order decomposition of
NO2(g) if the rate constant is 0.544 M-1 s-1 at a particular
temperature; the initial concentration of NO2 = 0.0100 M.
t1/2 =1
k2[A]o=
1
(0.544 𝑀−1 𝑠−1)(1.00 x 10−2 𝑀)
t1/2 = 184 s
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Zero Order Reactions
rate = -∆[A]t
∆t
In zero-order reactions, the
concentration of the
reactant has no effect on
the rate. One important
application is measuring
enzyme kinetics.
A (+ other reactants) → products
= k0[A]0 = k0
And the integrated rate law is -
[A] t
y = m x + b
m = -ko
[A]t = −kot + [A]o
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Half Life for Zero Order Reactions
When t = t½ ,
then [A]t1/2 = [A]o/2[A]t = −kot + [A]o
[A]o2
= −kot½ + [A]o
[A]o2
− [A]o = −kot½
−[A]o
2= −kot½
[A]o2ko
= t½
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Glucose bound to hexokinase
http://www.rcsb.org/pdb/explore/jmol.do?structureId=1BDG&bionumber=1
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Rate of product formation vs. substrate
concentration in an enzyme-catalyzed reaction.
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Summary of the Different Rate Laws and
How to Experimentally Distinguish Them
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• 13.1 Cars, Trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
• 13.4 Reaction Rates, Temperature, and the Arrhenius
Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Factors Affecting Rate
• Temperature:
o Increased temperature increases kinetic energy of molecules, molecular collisions
• Proper Molecular Orientation
• Activation Energy (Ea):
oMinimum energy of molecular collisions required to break bonds in reactants, leading to formation of products
• These factors are incorporated into the Arrhenius equation
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Effect of Temperature
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Molecular Orientation
O3 + NO → O2 + NO2
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Reaction Energy Profile
High-energy transition state (“activated complex”)
Ea =
Activation
energy
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The Transition State or “Activated Complex”
H2(g) + I2(g) → 2 HI(g) H = -13 kJ
H2 + I2
2 HI
I I
H H
Transition state or
activated complex
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Transition state - bonds between reactants are
breaking at the same time that bonds are
forming between the products
H - H
I - I→ →
H H
I I
H H
I I
Bonds breaking
Bonds forming
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Arrhenius Equation
• k = rate constant
• A is a “collisional frequency factor” , includes the probability of colliding with the correct orientation
• Ea = activation energy
• R = Ideal Gas Law constant = 8.314 J/mol K
• T = temperature (in kelvin)
All of these controlling factors are combined
together into the Arrhenius Equation…..
k = Ae−Ea/RT
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Ea and A can be measured graphically using
linearization again….
k = Ae-Ea/RT
ln k = ln (A▪e-Ea/RT) and since ln (a▪b) = ln a + ln b
ln k = ln A + ln (e-Ea/RT)
y = m x + b
ln k = −EaR
1T
+ ln A m = -Ea/R
b = ln A
ln k = ln A + - Ea/RT
Arrhenius Equation
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Graphical Determination of Ea
= +
- 1ln lnaEk A
R T
m = -Ea/R
b = ln A
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y = -1580x + 27.99
m = −EaR
Ea = −(8.314 J/mol K)(−1580 K) = 13,140 J = 13.1 kJ
Ea = −Rm
b = ln A A = eb
A = eb = e27.99 = 1.45 x 1012 s-1
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Measuring Ea using data from two
different temperatures -
= +
1
1
- 1ln lnaEk A
R T
= +
2
2
- 1ln lnaEk A
R T
Another way to measure k if only two data points are available
ln k1 − ln k2=−Ea
R
1
T1+ ln A −
−Ea
R
1
T2+ ln A
lnk1k2
=−Ea
R
1
T1−
1
T2ln
k1k2
=Ea
R
1
T2−
1
T1
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Back to Exercise
13.9 where the first
two experiments
have been chosen.
lnk1k2
=Ea
R
1
T2−
1
T1
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• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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Method of Initial
Rates: collect kinetic
data in the lab for
each reactant
General Rate Law:
analyze the kinetic data in
order to determine the
exponents on the
reactants
Propose a reasonable
reaction mechanism
that is consistent with
the rate law
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Reaction Mechanisms
• Reaction Mechanism: A set of steps that describe
how a reaction occurs at the molecular level; must be
consistent with the rate law for the reaction
• Elementary Step: Molecular-level view of a single
process taking place in a chemical reaction
• Intermediate: Species produced in one step of a
reaction and consumed in a subsequent step
• Molecularity: The number of ions, atoms, or
molecules involved in an elementary step in a reaction
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Most reactions involve a
sequence of steps
Overall Br2(g) + 2 NO(g) → 2 BrNO(g)
Reaction“intermediate”
Step 1 Br2(g) + NO(g) → Br2NO(g)
Step 2 Br2NO(g) + NO(g) → 2 BrNO(g)
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Each step in the mechanism is called an “elementary
step” and has it’s own Ea and rate constant, k.
Br2(g) + NO(g)Br2NO(g)
[+ NO(g)]
2 BrNO(g)
Ea1
Ea2
Reaction coordinate →
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Molecularity of Elementary Steps
1 molecule = unimolecular
2 molecules = bimolecular
3 molecules = termolecular
2. The exponents in the rate expressions for
elementary steps are the same as the
stoichiometric coefficients in each
elementary step. The reason why can be
explained by returning to Sec. 13-3…..
1. termolecular (and higher) reactions are
improbable because it's difficult to have 3 or more
molecules colliding simultaneously with the
correct energy and orientation.
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Why the rate is proportional to the product of
concentrations (Sec.13-3 revisited)
NO(g) + O3(g) → NO2(g) + O2(g) rate = k[NO][O3]
1 x 1 = 1 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 3 x 3 = 9
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What if the rate law contains exponents?
2 NO(g) + O2(g) → 2 NO2(g) rate = k[NO]2[O2]
When holding O2 constant, doubling NO
should increase the rate by 22 = 4X;
tripling should increase by 32 = 9X, etc.
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equivalent
equivalent
Doubling Concentrations
Tripling Concentrationsequivalent equivalent
equivalent equivalent
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equivalent
equivalent
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Reaction mechanisms have a RATE DETERMINING
STEP - the slowest step in the mechanism.
2 NH3 + OCl- → N2H4 + Cl- + H2O
Step 1 NH3 + OCl- → NH2Cl + OH- fast
Step 3 N2H5+ + OH- → N2H4 + H2O fast
Overall
Reaction2 NH3 + OCl- → N2H4 + Cl- + H2O
Step 2 NH2Cl + NH3 → N2H5+ + Cl- slow
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The overall rate expression must be
consistent with the rate law for the slow step
so the rate law = k[NH2Cl][NH3]
2 NH3 + OCl- → N2H4 + Cl- + H2O
Step 1 NH3 + OCl- → NH2Cl + OH- fast
Step 3 N2H5+ + OH- → N2H4 + H2O fast
Overall
Reaction2 NH3 + OCl- → N2H4 + Cl- + H2O
Step 2 NH2Cl + NH3 → N2H5+ + Cl- slow
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• The sum of the elementary steps must give the overall
balanced equation for the reaction.
• The rate-determining step is the slowest step in the
mechanism
• The rate law for the reaction is given by the slow step,
unless…
• There is a fast equilibrium step prior to the slow step, and so
a calculation has to be made starting with ratef = rater
(coming up)
• the detection of an intermediate supports a given
mechanism
Writing plausible reaction mechanisms:
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Example Reaction Mechanisms
2 NO2 (g) → 2 NO(g) + O2(g)
Proposed mechanism:
Step 1: 2 NO2(g) → NO(g) + NO3(g)
Step 2: NO3(g) → NO(g) + O2(g)
rate = k[NO2]2
Since the observed rate is k[NO2]2, the first
step must be the RDS.
Known from exp.
2 NO2 (g) → 2 NO(g) + O2(g)
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Rate Determining Step for Mechanisms
With an Equilibrium Elementary Step
2 NO(g) + O2(g) → 2NO2(g) rate = k[NO]2[O2]
Proposed mechanism:
Step 1: NO + O2 NO3 rate1 = k1[NO][O2] fast & rev.
Step 2: NO3 + NO → 2NO2 rate2 = k2[NO3][NO]
2 NO(g) + O2(g) → 2NO2(g)
Is this mechanism consistent
with the observed rate law?
© 2014 W. W. Norton Co., Inc.
Step 1: NO + O2 NO3
2 NO(g) + O2(g) → 2 NO2(g)
at equil. ratef = rater
ratef = kf [NO][O2]
kf
rater = kr [NO3]
kr
so kf [NO][O2] = kr [NO3]
by rearrangement: [NO3] =kr
[NO][O2]kf
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[NO3] =kr
[NO][O2]kf
= k’[NO]2[O2]
Substituting into rate2 = k2[NO3][NO]
= k2kr
[NO][O2]kf [NO]
• Which is consistent with
the observed rate law
• mechanism verified
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2 NO(g) + O2(g) → 2 NO2(g)
Step 1: NO + O2 NO3 Step 2: NO3 + NO → 2NO2
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© 2014 W. W. Norton Co., Inc.
HW# 107
At a given temperature, the rate law of the reaction between
NO and Cl2 is proportional to the product of concentrations of
the two gases: [NO][Cl2]. The following two-step mechanism
has been proposed for the reaction:
Step 1: NO(g) + Cl2(g) → NOCl2(g)
Step 2: NOCl2(g) + NO(g) → 2 NOCl(g)
Overall: 2 NO(g) + Cl2(g) → NOCl2(g)
Which step must be the RDS if this mechanism is correct?
© 2014 W. W. Norton Co., Inc.
HW# 108
Mechanism of Ozone Destruction:
Ozone decomposes thermally to oxygen in the following
mechanism:
O3(g) → O(g) + O2(g)
O(g) + O3(g) → 2 O2(g)
2 O3(g) → 3 O2(g)
The reaction is second order in ozone. What properties of the
two elementary steps (specifically, relative rate and
reversibility) are consistent with this mechanism?
Step 1:
Step 2:
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2 NO2(g) → N2O4(g)
N2O4(g) → N2O3(g) + O(g)
N2O3(g) + O(g) → N2O2(g) + O2(g)
N2O2(g) → 2 NO(g)
(a)slow
fast
fast
fast
HW# 110
The rate laws for the thermal and photochemical decomposition
of NO are different. Which of the following mechanisms are
possible for the thermal decomposition of NO2, and which are
possible for the photochemical decomposition of NO2? For
thermal decomposition, Rate = kT[NO2]2, and for photochemical
decomposition, Rate = kP[NO2].
© 2014 W. W. Norton Co., Inc.
2 NO2(g) → NO(g) + NO3(g)
NO3(g) → NO(g) + O2(g)
slow
fast
(b)
HW# 110, cont’d
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HW# 110, cont’d
NO2(g) → N(g) + O2(g)
N(g) + NO2(g) → N2O2(g)
N2O2(g) → 2 NO(g)
slow
fast
slow
(c)
© 2014 W. W. Norton Co., Inc.
• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction
Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
Chapter Outline
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© 2014 W. W. Norton Co., Inc.
Catalysts and the Ozone Layer
© 2014 W. W. Norton Co., Inc.
Ea for Step 2 = 17.7 kJ/mol (slow step)
Natural Photodecomposition
of Ozone:
Mechanism:
⎯⎯⎯→sunlight
3 2O ( ) O ( ) + O( )g g gStep 1:
⎯⎯→3 2O ( ) + O( ) 2O ( ) g g gStep 2:
2 O3(g) → 3 O2(g)Overall:
slow
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CFC Emissions and Ozone
→3 2ClO( ) + O ( ) Cl( ) + O ( )g g g g
Ea for Cl-catalyzed reaction = 2.2 kJ/mol
Chlorofluorocarbons (CFC) e.g. CCl2F2, CCl3F, CClF3
Stable! Migrate into the stratosphere
Net: 2 O3(g) → 3 O2(g)
⎯⎯→hν
3 2CCl F( ) CCl F( ) + Cl( )g g g
→3 2Cl( ) + O ( ) ClO( ) + O ( )g g g g
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Energy Profiles for O3 Decomposition
⎯⎯→ν
3 22 O 3 Oh
Cl atom:
• Not consumed
during the reaction
• Homogeneous
catalyst
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Types of Catalysts
Homogeneous
Catalyst:• One in the same
phase as the
reactants
Heterogeneous
Catalyst:• One in a different
phase from the
reactants
Platinum-rhodium gauze
catalyst used in the
production of HNO3
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Automobile Catalytic Converter
Catalyst = Pt-NiO, finely divided powder. Higher
surface area provides more sites for adsorption and
reaction
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Catalytic Converters
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The heterogeneous catalyst provides a surface where
the reactants can be ADSORBED on an ACTIVE SITE -
3 H2(g) + N2(g) →
2 NH3(g)
Fe
Industrial production
of ammonia (Haber
Process)-
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