© 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2 · © 2008 Brooks/Cole 1 Chapter 13: Chemical Kinetics:...
Transcript of © 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2 · © 2008 Brooks/Cole 1 Chapter 13: Chemical Kinetics:...
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© 2008 Brooks/Cole 1
Chapter 13: Chemical Kinetics: Rates of
Reactions
© 2008 Brooks/Cole 2
Chemical Kinetics
“The study of speeds of reactions and the
nanoscale pathways or rearrangements by which
atoms and molecules are transformed to products”
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Reaction Rate
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Reaction Rate
Combustion of Fe(s) powder:
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Reaction Rate
Change in [reactant] or [product] per unit time.
rate = = change in concentration of Cv+
elapsed time
[Cv+ ]
t
Cresol violet (Cv+; a dye) decomposes in NaOH(aq):
Cv+(aq) + OH-(aq) CvOH(aq)
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Time, t [Cv+] Average rate
(s) (mol / L) (mol L-1 s-1)
0.0 5.000 x 10-5
10.0 3.680 x 10-5
20.0 2.710 x 10-5
30.0 1.990 x 10-5
40.0 1.460 x 10-5
50.0 1.078 x 10-5
60.0 0.793 x 10-5
80.0 0.429 x 10-5
100.0 0.232 x 10-5
13.2 x 10-7
9.70 x 10-7
7.20 x 10-7
5.30 x 10-7
3.82 x 10-7
2.85 x 10-7
1.82 x 10-7
0.99 x 10-7
Average rate of the Cv+ reaction can be calculated:
Reaction Rate
2
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Reaction Rates and Stoichiometry
Stoichiometry:
Loss of 1 Cv+ Gain of 1 CvOH
Rate of Cv+ loss = Rate of CvOH gain
Another example:
2 N2O5(g) 4 NO2(g) + O2(g)
Negative rate Positive rate
Rate of loss of N2O5 divided by -2, equals rate of gain of O2
Cv+(aq) + OH-(aq) CvOH(aq)
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For any general reaction:
a A + b B c C + d D
The overall rate of reaction is:
Reactants decrease with time.
Negative sign. Products increase with time.
Positive sign
Reaction Rates and Stoichiometry
Rate = = = = 1
a
[A]
t 1
b
[B]
t +
1
c
[C]
t +
1
d
[D]
t
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Reaction Rates and Stoichiometry
H2 (g) + I2 (g) 2 HI (g)
the rate of loss of I2 is 0.0040 mol L-1 s-1. What is
the rate of formation of HI ?
Rate = = = [H2]
t
[I2]
t +
1
2
[HI]
t
Rate = = (-0.0040) = [H2]
t +
1
2
[HI]
t
For:
[HI]
t So = +0.0080 mol L-1 s-1
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Average Rate and Instantaneous Rate
5.0E-5
4.0E-5
3.0E-5
2.0E-5
1.0E-5
0
[Cv
+]
(m
ol/L)
0 20 40 60 80 100 t (s)
Graphical view of Cv+ reaction:
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Average Rate and Instantaneous Rate
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Rate may change when [reactant] changes.
t [Cv+] Rate of Cv+ Rate/[Cv+]
(s) (M) loss (M / s) (s-1)
0 5.00 x 10-5 1.54 x 10-6 0.0308
80 4.29 x 10-6 1.32 x 10-7 0.0308
• Cv+ example shows this.
• For Cv+ the rate is proportional to concentration.
Effect of Concentration on Reaction Rate
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Rate Law and Order of Reaction
A general reaction will usually have a rate law:
rate = k [A]m [B]n . . .
The orders are usually integers (-2, -1, 0, 1, 2…),
but may also be fractions ( , …)
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Determining Rate Laws from Initial Rates
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Data for the reaction of methyl acetate with base:
CH3COOCH3 + OH- CH3COO- + CH3OH
Rate law:
rate = k [CH3COOCH3]m [OH-]n
Determining Rate Laws from Initial Rates
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Dividing the first two data sets:
4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n
2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n
1 raised to any
power = 1
Determining Rate Laws from Initial Rates
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Use experiments 2 & 3 to find m:
9.0 x 10-4 M/s = k (0.080 M)m(0.080 M)n
4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n
Determining Rate Laws from Initial Rates
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Determining Rate Laws from Initial Rates
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If a rate law is known, k can be determined:
2.2 x 10-4 M/s
(0.040 M)(0.040 M) k =
k = 0.1375 M-1 s-1 = 0.1375 L mol-1 s-1
rate
[CH3COOCH3 ][OH-] k =
Could repeat for each run, take an average…
But a graphical method is better.
Using run 1:
Determining Rate Laws from Initial Rates
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The Integrated Rate Law
rate = – = k [A] [A]
t
(as a differential equation) = – = k
[A]
d [A]
dt
Calculus is used to integrate a rate law.
Consider a 1st-order reaction: A
products
Integrates to: ln [A]t = k t + ln [A]0
y = m x + b (straight line)
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Order Rate law Integrated rate law Slope
The Integrated Rate Law
The reaction:
0 rate = k [A]t = -kt + [A]0 -k
1 rate = k[A] ln[A]t = -kt + ln[A]0 -k
A
products doesn’t have to be 1st order.
Some common integrated rate laws:
1
[A]t 2 rate = k[A]2 = kt + +k
1
[A]0
y x
The most accurate k is obtained from the slope of a plot.
ln[A
] time t
First-order reaction
slope = -k
1/[
A]
time t
Second-order reaction
slope = k
[A]
time t
Zeroth-order reaction
slope = -k
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The Integrated Rate Law
• The reaction is first order (the only linear plot)
• k = -1 x (slope) of this plot.
Rate data for the decomposition of cyclopentene
C5H8(g) C5H6(g) + H2(g)
were measured at 850°C. Determine the order of the
reaction from the following plots of those data:
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Half-lives are only useful for 1st -order reactions.
Why?
Half-Life
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For a 1st-order reaction:
ln[A]t = -kt + ln[A]0
Half-Life
When t = t1/2 [A]t = [A]0
Then: ln( [A]0) = -kt1/2 + ln[A]0
ln( [A]0/[A]0) = -kt1/2 {note: ln x – ln y = ln(x/y)}
ln( ) = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y }
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Half Life
0.010
0.008
0.006
0.004
0.002
0
[cis
pla
tin]
(m
ol/L)
0 400 800 1200 1600 2000 t (min)
• the cisplatin lost after 475 min. • (0.0100 M 0.0050 M)
• [cisplatin] halves every 475 min
t1/2 of a 1st-order reaction can be used to find k.
For cisplatin (a chemotherapy agent):
k = = 0.693
475 min ln 2
t1/2
= 1.46 x 10-3 min-1
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Use an integrated rate equation.
a) [reactant],1600.s after initiation.
b) t for [reactant] to drop to 1/16th of its initial value.
c) t for [reactant] to drop to 0.0500 mol/L.
Calculating [ ] or t from a Rate Law
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In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s
(a) Calculate [reactant] ,1600.s after initiation.
1st order:
k = ln 2/ t = 0.6931/(400. s) = 1.733x10-3 s-1
and ln [A]t = -kt + ln [A]0
so ln[A]t = -(0.001733 s-1)(1600 s) +ln(0.500)
ln[A]t = -2.773 + -0.693 = -3.466
[A]t = e-3.466 = 0.0312 mol/L
Calculating [ ] or t from a Rate Law
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In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s
(b) Calculate t for [reactant] to drop to 1/16th of its initial value.
Note: part (a) could be solved in a similar way. 1600 s = 4 t1/2
so 0.500 0.250 0.125 0.0625 0.0313 M.
[reactant]0 [reactant]0 t1/2 1
2
[reactant]0 [reactant]0 t1/2 1
4
1
2
[reactant]0 [reactant]0 t1/2 1
8
1
4
[reactant]0 [reactant]0 t1/2 1
16
1
8
4 t1/2 = 4 (400 s) = 1600 s
Calculating [ ] or t from a Rate Law
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From part (a): k = 1.733 x 10-3 s-1
ln [A]t = -kt + ln [A]0
then ln (0.0500) = -(0.001733 s-1) t + ln(0.500)
-2.996 = -(0.001733 s-1) t – 0.693
t = 1.33 x 103 s
In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s
(c) Calculate t for [reactant] to drop to 0.0500 mol/L ?
t = -2.303
-0.001733 s-1
Calculating [ ] or t from a Rate Law
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Nanoscale View: Elementary Reactions
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Elementary reactions
unimolecular
bimolecular
unimolecular
unimolecular
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Unimolecular Reactions
2-butene isomerization is unimolecular:
C=C
CH3
H H
H3C (g) C=C
CH3
H
H
H3C (g)
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Final state Initial state
Reaction Progress (angle of twist)
E = -7 x 10-21 J
Ea =
435 x
10
-21 J
500
400
300
200
100
0 Pote
ntial energ
y (
10
-21 J
)
-30° 0 30° 60° 90° 120° 150° 180° 210°
cis-trans conversion twists the C=C bond.
• This requires a lot of energy (Ea= 4.35x10-19J/molecule = 262 kJ/mol)
• Even more (4.42x10-19J/molecule) to convert back.
transition state or
activated complex
Ea is the activation
energy, the minimum E to
go over the barrier.
Exothermic overall
Unimolecular Reactions
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Transition State
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Bimolecular Reactions
I- must collide with enough E and in the right location
to cause the inversion.
e.g. Iodide ions reacting with methyl bromide:
I-(aq) + CH3Br(aq) ICH3(aq) + Br-(aq)
transition state
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Bimolecular Reactions
I- must collide in the right
location to cause the inversion.
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• Also has an activation barrier (Ea).
• Forward and back Ea are different.
• Here the forward reaction is endothermic.
Products
(final state)
Reactants
(initial state)
Reaction Progress (changing bond lengths and angles)
Ea =
126 x
10
-21 J
E = 63 x 10-21 J
150
120
90
60
30
0 Pote
ntial energ
y (
10
-21 J
)
transition state
Bimolecular Reactions
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Temperature and Reaction Rate
Increasing T will speed up most reactions.
Higher T = higher average Ek for the reactants.
= larger fraction of the molecules can
overcome the activation barrier.
25°C
kinetic energy
nu
mb
er
of
mo
lecu
les
Ea
75°C Many more molecules have
enough E to react at 75°C,
so the reaction goes much faster.
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Temperature and Reaction Rate
Reaction rates are strongly T-dependent. Data for
the I- + CH3Br reaction:
250 300 350 400 T (K)
0.0
0
0
.10
0.2
0
0.3
0
k (
L m
ol-1
K-1
)
T (K) k (L mol-1 K-1)
273 4.18 x 10-5
290 2.00 x 10-4
310 2.31 x 10-3
330 1.39 x 10-2
350 6.80 x 10-2
370 2.81 x 10-1
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-Ea / RT k = A e
Quantity Name Interpretation and/or comments
A Frequency factor How often a collision occurs with
the correct orientation.
Ea Activation energy Barrier height.
e-Ea/RT Fraction of the molecules with enough E to cross the barrier.
T Temperature Must be in kelvins.
R Gas law constant 8.314 J K-1 mol-1.
Temperature and Reaction Rate
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Determining Activation Energy
Take the natural logarithms of both sides:
-Ea / RT ln k = ln A e
1
T ln k = + ln A
Ea
R
A plot of ln k vs. 1/T is linear (slope = Ea/R).
-Ea / RT ln k = ln A + ln e
ln k = ln A + ln e Ea
RT
ln e = 1
ln ab = ln a + ln b
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Determining Activation Energy
The iodide-methyl bromide reaction data:
intercept = 23.85
slope = -9.29 x 10 3 K
0 0.001 0.002 0.003 0.004
28
18
8
-2
-12
ln k
1/T (K-1)
Ea = -(slope) x R
= -(-9.29 x103 K) 8.314 J
K mol
= 77.2 x 103 J/mol
= 77.2 kJ/mol
A = eintercept = e23.85
A = 2.28 x 1010 L mol-1 s-1
(A has the same units as k)
8
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Rate Laws for Elementary Reactions
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Reaction Mechanisms
When [H3O+] is between 10-3 M and 10-5 M,
rate = k [I-][H2O2]
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Reaction Mechanisms
2 I-(aq) + H2O2(aq) + 2 H3O+(aq) I2 (aq) + 4 H2O(l)
slow
fast
fast
overall 2 I- + H2O2 + 2 H3O+ I2 + 4 H2O
Shows the
bonding in H2O2
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Reaction Mechanisms
© 2008 Brooks/Cole 47
Reaction Mechanisms
A good analogy is supermarket shopping:
• You run in for 1 item (~1 min = fast step), but…
• The checkout line is long (~10 min = slow step).
• Time spent is dominated by the checkout-line wait.
• In a reaction, a slow step may be thousands or
even millions of times slower than a fast step.
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Reaction Mechanisms
The overall rate is expected to be
rate = k [H2O2][ I- ] as observed!
9
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Consider:
2 NO (g) + Br2 (g) 2 NOBr (g)
Mechanisms with a Fast Initial Step
The generally accepted reaction mechanism is:
2 NO + Br2 2 NOBr
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Mechanisms with a Fast Initial Step
NO + Br2 NOBr2 fast, reversible
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Mechanisms with a Fast Initial Step
NO + Br2 NOBr2 reversible, equilibrium k1
k-1
rate forward = rate back
k1[NO][Br2] = k-1[NOBr2]
[NOBr2] = k1[NO][Br2]
k-1
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Mechanisms with a Fast Initial Step
rate = k2 [NOBr2] [NO]
The earlier rate law:
k1[NO][Br2]
k-1 rate = k2 [NO]
k1k2
k-1 rate = [Br2][NO]2
Now only contains starting materials - can be
checked against experiment.
becomes:
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Summary
Elementary reactions: the rate law can be written
down from the stoichiometry.
unimolecular rate = k[A]
bimolecular rate = k[A]2 or
rate = k[A][B]
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Catalysts and Reaction Rate
10
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Catalysts and Reaction Rate
2-butene isomerization is catalyzed by a trace of I2.
No catalyst:
rate = k [cis-2-butene]
A trace of I2(g) speeds up the reaction, and:
rate = k [ I2 ] [cis-2-butene]
C=C
CH3
H H
H3C
(g) C=C CH3
H
H
H3C
(g)
k uncatalyzed k
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I2 splits into 2 atoms.
Each has an unpaired
e-. (shown by the dot)
I• attaches and breaks
one C-C bond
C=C
CH3
H H
H3C
C–C H
CH3
H
H3C
I •
I2 is not in the overall equation, and is not used up.
The mechanism changes!
Catalysts and Reaction Rate
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Loss of I• and formation of C=C
C–C CH3
H
H
H3C
I •
C=C CH3
H
H
H3C
+ I•
C–C H
CH3
H
H3C
I •
C–C CH3
H
H
H3C
I •
Rotation around C-C
I2 is regenerated
Catalysts and Reaction Rate
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Products
(final state)
Reactants
(initial state)
Reaction Progress
E = -4 kJ/mol
Ea = 115 kJ/mol
I• leaves; double bond
reforms I2 dissociates to
I• + I•
Transition state for the
uncatalyzed reaction
I• + I• regenerates I2
I• adds to cis-2-butene,
(double single bond)
Rotation around C-C
Ea = 262 kJ/mol
Catalysts and Reaction Rate
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Catalysts and Reaction Rate
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Enzymes: Biological Catalysts
11
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Enzyme Activity and Specificity
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Enzyme Activity and Specificity
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Enzyme Activity and Specificity
Enzymes are effective catalysts because they:
• Bring and hold substrates together while a reaction occurs.
• Hold substrates in the shape that is most effective for
reaction.
• Can donate or accept H+ from the substrate (act as acid or
base)
• Stretch and bend substrate bonds in the induced fit so the
reaction starts partway up the activation-energy hill.
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Enzyme kinetics
Products (final state)
Reactants
(initial state)
Reaction Progress
E
Ea
E'a
Transformation of
the substrate to
products
Activation energy E'a is
much smaller than Ea
and so the enzyme
makes the reaction
much faster
Pote
ntia
l energ
y, E
Formation of the
enzyme-substrate
complex
Transition state for the
uncatalyzed reaction
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Enzyme Activity and Specificity
Enzyme catalyzed reactions:
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Catalysis in Industry
CH3OH(l) + CO(g) CH3COOH(l) RhI3
auto exhausts are cleaned by catalytic converters:
2 CO(g) + O2(g) 2 CO2
2 C8H18(g) + 25 O2 16 CO2(g) + 18 H2O(g)
2 NO(g) N2(g) + O2(g)
Pt-NiO
Pt-NiO
catalyst
12
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…they form N2 and O2 and
leave the surface.
Controlling Automobile Emissions
…N and O migrate
on the surface until they get close to like
atoms…
…dissociates into N
and O atoms (each bonded to Pt)…
…forms a bond with
the Pt surface…
NO approaches
the Pt surface…
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Converting Methane to Liquid Fuel
Methane is hard to transport. It can be converted
to methanol:
CH4(g) + O2(g) CO(g) + 2 H2(g)
CO(g) + 2 H2(g) CH3OH(l)
A Pt-coated ceramic catalyst
allows the 1st reaction to occur
at low T.