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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”
© 2008 Brooks/Cole 3
Reaction Rate
© 2008 Brooks/Cole 4
Reaction Rate Combustion of Fe(s) powder:
© 2008 Brooks/Cole 5
Reaction Rate
Change in [reactant] or [product] per unit time.
rate = = change in concentration of Cv+
elapsed time Δ [Cv+ ] Δt
Cresol violet (Cv+; a dye) decomposes in NaOH(aq):
Cv+(aq) + OH-(aq) → CvOH(aq)
© 2008 Brooks/Cole 6
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
© 2008 Brooks/Cole 7
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)
© 2008 Brooks/Cole 8
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 Δ[A] Δt − 1
b Δ[B] Δt + 1
c Δ[C] Δt
+ 1 d Δ[D] Δt
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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 = = = − Δ[H2] Δt − Δ[I2]
Δt + 1 2 Δ[HI] Δt
Rate = − = − (-0.0040) = Δ[H2] Δt
+ 1 2 Δ[HI] Δt
For:
Δ[HI] Δt
So = +0.0080 mol L-1 s-1
© 2008 Brooks/Cole 10
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
3
© 2008 Brooks/Cole 13
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 (½, ⅓…)
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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
© 2008 Brooks/Cole 16
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
© 2008 Brooks/Cole 17
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
4
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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
© 2008 Brooks/Cole 20
The Integrated Rate Law
rate = – = k [A] Δ[A] Δt
(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)
© 2008 Brooks/Cole 21
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
© 2008 Brooks/Cole 22
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:
© 2008 Brooks/Cole 23
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 = ½[A]0
Then: ln(½[A]0) = -kt1/2 + ln[A]0
ln(½[A]0/[A]0) = -kt1/2 {note: ln x – ln y = ln(x/y)} ln(½) = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y }
5
© 2008 Brooks/Cole 25
Half Life
0.010
0.008
0.006
0.004
0.002
0
[cis
plat
in]
(mol
/L)
0 400 800 1200 1600 2000 t (min)
• ½ 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
© 2008 Brooks/Cole 26
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
© 2008 Brooks/Cole 27
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/ t½ = 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
© 2008 Brooks/Cole 30
Nanoscale View: Elementary Reactions
6
© 2008 Brooks/Cole 31
Elementary reactions
unimolecular
bimolecular
unimolecular
unimolecular
© 2008 Brooks/Cole 32
Unimolecular Reactions 2-butene isomerization is unimolecular:
C=C CH3
H H
H3C (g) C=C
CH3
H
H
H3C (g)
© 2008 Brooks/Cole 33
Final state Initial state
Reaction Progress (angle of twist)
ΔE = -7 x 10-21 J
E a =
435
x 1
0-21
J
500
400
300
200
100
0 Pot
entia
l ene
rgy
(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
© 2008 Brooks/Cole 35
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
© 2008 Brooks/Cole 36
Bimolecular Reactions
I- must collide in the right location to cause the inversion.
7
© 2008 Brooks/Cole 37
• 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)
E a =
126
x 1
0-21
J
ΔE = 63 x 10-21 J
150
120
90
60
30
0 Pot
entia
l ene
rgy
(10-
21 J
)
transition state
Bimolecular Reactions
© 2008 Brooks/Cole 38
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
num
ber o
f mol
ecul
es
Ea
75°C Many more molecules have enough E to react at 75°C, so the reaction goes much faster.
© 2008 Brooks/Cole 39
Temperature and Reaction Rate Reaction rates are strongly T-dependent. Data for the I- + CH3Br reaction:
250 300 350 400 T (K)
0.00
0.1
0
0
.20
0.30
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
© 2008 Brooks/Cole 40
-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
© 2008 Brooks/Cole 41
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
© 2008 Brooks/Cole 42
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]
© 2008 Brooks/Cole 45
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
© 2008 Brooks/Cole 46
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.
© 2008 Brooks/Cole 48
Reaction Mechanisms
The overall rate is expected to be rate = k [H2O2][ I- ] as observed!
9
© 2008 Brooks/Cole 49
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
© 2008 Brooks/Cole 50
Mechanisms with a Fast Initial Step
NO + Br2 NOBr2 fast, reversible
© 2008 Brooks/Cole 51
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
© 2008 Brooks/Cole 52
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:
© 2008 Brooks/Cole 53
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]
© 2008 Brooks/Cole 54
Catalysts and Reaction Rate
10
© 2008 Brooks/Cole 55
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 ]½ [cis-2-butene]
C=C CH3
H H
H3C (g) C=C
CH3
H
H
H3C (g)
k ≠ uncatalyzed k
© 2008 Brooks/Cole 56
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
© 2008 Brooks/Cole 57
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
© 2008 Brooks/Cole 58
Products (final state)
Reactants (initial state)
Reaction Progress
ΔE = -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
© 2008 Brooks/Cole 59
Catalysts and Reaction Rate
© 2008 Brooks/Cole 60
Enzymes: Biological Catalysts
11
© 2008 Brooks/Cole 61
Enzyme Activity and Specificity
© 2008 Brooks/Cole 62
Enzyme Activity and Specificity
© 2008 Brooks/Cole 63
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.
© 2008 Brooks/Cole 64
Enzyme kinetics
Products (final state)
Reactants (initial state)
Reaction Progress
ΔE
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
Potential energy, E
Formation of the enzyme-substrate
complex
Transition state for the uncatalyzed reaction
© 2008 Brooks/Cole 65
Enzyme Activity and Specificity Enzyme catalyzed reactions:
© 2008 Brooks/Cole 66
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
© 2008 Brooks/Cole 67
…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…
© 2008 Brooks/Cole 68
Converting Methane to Liquid Fuel Methane is hard to transport. It can be converted to methanol:
CH4(g) + ½ 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.