Transcript 13 Chemical Kinetics - University of Waterloo

15 Chemical Kinetics
Chemical kinetics: the study of reaction rate,  a quantity
conditions affecting it,
the molecular events during a chemical reaction (mechanism), and
presence of other components (catalysis).
Factors affecting reaction rate:
Concentrations of reactants
Catalyst
Temperature
Surface area of solid reactants or catalyst
What quantities do we study regarding chemical reactions?
15 Chemical Kinetics
1
Reaction Rate Defined
Reaction rate: changes in a concentration
of a product or a reactant per unit time.
[ ]
Reaction rate = ——
t
[]
concentration
change
[ ]
t
Define reaction rate and explain
Average reaction rate
Instantaneous reaction rate
(2 tangents shown)
Initial reaction rate
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t
2
Expressing reaction rates
For a chemical reaction, there are many ways to express the reaction
rate. The relationships among expressions depend on the equation.
Note the expression and reasons for their relations for the reaction
2 NO + O2 (g) = 2 NO2 (g)
[O2]
1 [NO]
1 [NO2]
Reaction rate = – ——— = – — ———— = — ———
t
2
t
2 t
Make sure you can write expressions for any reaction and figure out
the relationships. For example, give the reaction rate expressions for
2 N2O5 = 4 NO2 + O2
How can the rate expression be15 unique
and universal?
Chemical Kinetics
3
Calculating reaction rate
The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and
1200 s after the reactants are mixed at the appropriate temperature.
Evaluate the reaction rates for
2 N2O5 = 4 NO2 + O2
Solution:
(0.93 – 1.24)e-2
– 0.31e-2 M
Decomposition rate of N2O5 = – ———————— = – ——————
1200 – 600
600 s
= 5.2e-6 M s-1.
Note however,
rate of formation of NO2 = 1.02e-5 M s-1.
rate of formation of O2 = 2.6e-6 M s-1.
Be able to do this type problems
15 Chemical Kinetics
The reaction rates are
expressed in 3 forms
4
Determine Reaction Rates
To measure reaction rate, we measure the concentration
of either a reactant or product at several time intervals.
The concentrations are measured using spectroscopic
method or pressure (for a gas). For example, the total
pressure increases for the reaction:
2 N2O5 (g)  4 NO2 (g) + O2(g)
barometer
Because 5 moles of gas products are produced from 2
moles of gas reactants. For the reaction
CaCO3 (s)  CaO(s) + CO2 (g)
The increase in gas pressure is entirely due to CO2
formed.
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5
Differential Rate Laws
Dependence of reaction rate on the concentrations of reactants is
called the rate law, which is unique for each reaction.
For a general reaction, a A + b B + c C  products
the rate law has the general form
order wrt A, B, and C, determined experimentally
reaction rate = k [A]X [B]Y [C]Z
the rate constant
For example, the rate law is
rate = k [Br-] [BrO3-] [H+]
for
5 Br- + BrO3- + 6 H+  3Br2 + 3 H2O
Use differentials
to express rates
The reaction is 1st order wrt all three reactants, total order 3.
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Variation of Reaction rates and Order
2nd order, rate = k [A]2
rate
First order, rate = k [A]
k = rate, 0th order
[A] = ___?
[A]
The variation of reaction rates as functions of concentration for
various order is interesting.
15 Chemical
Kinetics tool, worth noticing.
Mathematical analysis is an important
scientific
7
Differential Rate Law determination
Estimate the orders and rate constant k from the results observed for the
reaction? What is the rate when [H2O2] = [I-] = [H+] = 1.0 M?
H2O2 + 3 I- + 2 H+  I3- + 2 H2O
Exprmt
1
2
3
4
[H2O2]
0.010
0.020
0.010
0.010
[I-]
0.010
0.010
0.020
0.010
[H+]
0.0050
0.0050
0.0050
0.0100
Initial rate M s-1
1.15e-6
2.30e-6
2.30e-6
1.15e-6
Learn the strategy to determine the rate law from this example.
Figure out the answer without writing down anything.
Solution next
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Differential Rate Law determination - continue
Estimate the orders from the results observed for the reaction
H2O2 + 3 I- + 2 H+  I3- + 2 H2O
Exprmt
1
2
3
4
[H2O2]
0.010
0.020
0.010
0.010
[I-]
0.010
0.010
0.020
0.010
[H+]
0.0050
0.0050
0.0050
0.0100
Initial rate M s-1
1.15e-6
2.30e-6 1 for H2O2
2.30e-6 1 for I1.15e-6 0 for H+
1.15e-6 = k [H2O2]x [I-]y [H+]z
1.15e-6 k (0.010)x(0.010)y(0.0050)z
----------- = ------------------------------------2.30e-6 k (0.020)x(0.010)y(0.0050)z
x=1
 exprmt 1
 exprmt 2
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1
---- =
2
1
--- x
2
( )
9
Differential Rate Law determination - continue
Other orders are determined in a similar way as shown before.
Now, lets find k and the rate
Thurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1
k = 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1
And the rate law is therefore,
– d [H2O2]
k
rate = ————— = 0.0115 [H2O2] [I-]
dt
a differential rate law
total order 2
The rate when [H2O2] = [I-] = [H+] = 1.0 M:
The rate is the same as the rate constant k, when concentrations of
reactants are all unity (exactly 1), doesn’t matter what the orders are.
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10
Differential Rate Law determination – continue
The reaction rate –d[H2O2]/dt = 0.0115 [H2O2] [I–], for
H2O2 + 3 I- + 2 H+  I3- + 2 H2O
What is – d[I–]/dt when [H2O2] = [I–] = 0.5?
Solution: Please note the stoichiometry of equation and how the rate changes.
– d[I–]/dt = – 3 d[H2O2]/dt = 3* 0.0115 [H2O2] [I–]
= 0.0345 * 0.5 * 0.5
= 0.0086 M s-1
In order to get a unique rate constant k, we evaluate k for the
reaction
a A + b B  product
this way
rate = -1/a d[A]/dt = -1/b d[B]/dt = k [A]x [B]y
Chemical
Note the reaction rate expression15 and
theKinetics
stoichiometry of equation.
11
Differential Rate Law determination - continue
From the following reaction rates observed in 4 experiments, derive the
rate law for the reaction
A + B + C  products
where reaction rates are measured as soon as the reactants are mixed.
Expt
[A]o
[B]o
[C]o
rate
1
0.100
0.100
0.100
0.100
2
0.200
0.100
0.100
0.800
3
0.200
0.300
0.100
7.200
4
0.100
0.100
0.400
0.400
This example illustrates the strategy to determine, and a
reliable method to solve rate-law experimentally.
15 Chemical Kinetics
Solution next
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Differential Rate Law determination - continue
From the following reaction rates, derive the rate law for the reaction
A + B + C  products
where reaction rates are measured as soon as the reactants are mixed.
Expt
[A]o
[B]o
[C]o
rate
1
0.100
0.100
0.100
0.100
2
0.200
0.100
0.100
0.800
3
0.200
0.300
0.100
7.200
4
0.100
0.100
0.400
0.400
order
3 from expt 1 & 2
2 expt 1, 2 & 3
1 expt 1 & 4
Assume rate = k [A]x[B]y[C]z
Therefore 8 = 2x
log 8 = x log 2
x = log 8 / log 2
=3
0.800 k 0.2x 0.1y 0.1z
----- = ---------------------0.100 k 0.1x 0.1y 0.1z
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Integrated Rate Laws
concentrations as functions of time
One reactant A decomposes in 1st or 2nd order rate law.
Differential rate law
Integrated rate law
– d[A] / dt = k
[A] = [A]o – k t
d[A]
– —— = k [A]
dt
[A] = [A]o e – k t or ln [A] = ln [A]o – k t
d[A]
– —— = k [A]2
dt
1
1
—— – —— = k t
[A]
[A]o
[A] conc at t
[A]o conc at t = 0
Describe, derive and apply the integrated rate laws
Learn the strategy to determine rate-law
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Concentration and time of 1st order reaction
Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order
reactions. Apply the technique to evaluate k or [A] at various times.
[A]
ln[A]
ln [A] = ln [A]o – k t
[A] = [A]o e – k t
t½
t
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t
15
Half life & k of First Order Decomposition
The time required for half of A to decompose is called half life t1/2.
Since
[A] = [A]o e – k t
or
When
t = t1/2, [A] = ½ [A]o
Thus
ln ½ [A]o = ln [A]o – k t1/2
ln [A] = ln [A]o – k t
– ln 2 = – k t1/2
k t1/2 = ln 2 = 0.693  relationship between k and t1/2
Radioactive decay usually follow 1st order kinetics, and half life of an
isotope is used to indicate its stability.
Evaluate t½ from k or k from t½
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1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it
decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.
If the rate-law is known, what are the key parameters?
Solution next
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1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed
in 30 s. Estimate k, t½ and percent decomposed in 500 s.
Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or
0.9 = 1.0 e – k t apply [A]o = [A] e– k t
ln 0.9 = ln 1.0 – k 30 s
– 0.1054 = 0 – k * 30
k = 0.00351 s – 1
t½ = 0.693 / k = 197 s apply k t ½ = ln 2
[A] = 1.0 e – 0.00351*500 = 0.173
Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 %
After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.
After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.
Chemical
Apply integrated 15
rate
law Kinetics
to solve problems
18
Typical Problem wrt 1st Order Reaction
The decomposition of A is first order, and [A] is monitored. The following
data are recorded:
t / min 0
[A]/[M] 0.100
2
0.0905
4
0.0819
8
0.0670
Calculate k (What is the rate constant? k = 0.0499)
Calculate the half life (What is the half life? Half life = 13.89)
Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)
Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A
to decompose.)
Work out all the answers
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A 2nd Order Example
Dimerization of butadiene is second order:
2 C4H6(g) = C8H12(g).
The rate constant k at some temperature is 0.100 /min. The initial
concentration of butadiene [B] is 2.0 M.
Calculate the time required for [B] = 1.0 and 0.5 M
Calculate concentration of butadiene when t = 1, 5, 10, and 30.
If the rate-law is known, what are the key parameters?
Apply the right model and work out all the parameters.
15 Chemical Kinetics
Solution next
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A 2nd Order Example
1
1
—— – ——
[B]
[B]o
Dimerization of butadiene is second order:
2 C4H6(g) = C8H12(g).
1
1
—— – ——
[B]
[B]o
t = ————————
k
The rate constant k at some temperature
is 0.100 /min. The initial concentration of
butadiene [B] is 2.0 M.
Calculate the time t required for [B] = 1.0
and 0.5 M
Calculate concentration of butadiene
when t = 1, 5, 10, and 30.
t=
1
5
10
15
[B] = 1.67 1.0 0.67 0.50
Work out the formulas and
then evaluate values
=kt
30
35
0.29
0.25
15 Chemical Kinetics
[B]o
[B] = ——————
[B]o k t + 1
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Half life of 2nd Order Chemical Kinetics
4.5
4
3.5
3
1/[B]
2.5
2
1.5
1
0.5
0
t=
1
5
10
15
[B] = 1.67 1.0 0.67 0.50
30
35
0.29
0.25
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Kinetics
How does half life vary in 2nd order
reactions?
22
Plot of [B] vs. t & 1/[B] vs. t for 2nd Order Reactions
t=
1
5
10
15
[B] = 1.67 1.0 0.67 0.50
[B]
30
35
0.29
0.25
1
—
[B]
[B]o
[B] = ——————
[B]o k t + 1
t
What kind of plot is linear for
15 Chemical
Kinetics
1st and
2nd reactions?
1
——
[B]
1
– —— = k t
[B]o
t
23
Chemical Reaction and Molecular Collision
Molecular collisions lead to chemical reactions.
Thus, the reaction constant, k is determined by
several factors.
k=Zfp
Z: collision frequency
constant
p, the fraction with proper orientation
f, fraction of collision having sufficient energy for reaction
f is related to the potential energy barrier called
activation energy, Ea.
f  e – Ea / RT or exp (– Ea / R T)
Thus,
k = A e – Ea / RT
constant
Potential
energy
Ea
reaction
How does temperature affect reaction rates?
24
Explain energy aspect in a chemical reaction
15 Chemical Kinetics
Energy in chemical reactions
Potential
energy
R +A 
H
exothermic
RA-PD
activated
complex
Ea Ea for reverse reaction
 P+D
Endothermic rxn
Progress of reaction
15 Chemical
Kinetics in a chemical reaction
Explain the various terms and energy
changes
25
The Arrhenius Equation
The temperature dependence of the rate constant k is best
described by the Arrhenius equation:
or
k = A e – Ea / R T
ln k = ln A – Ea / R T
If k1 and k2 are the rate constants at T1 and T2 respectively, then
k1
Ea 1
1
ln —— = – — — – —
k2
R T1 T2
1903 Nobel Prize citation” …in recognition of the extraordinary services he has
rendered to the advancement of chemistry by his electrolytic theory of dissociation”
How does temperature affect reaction rates?
Derive and apply these relationship to solve problems,
15 Chemical Kinetics
and recall the Clausius-Clapeyron
equation.
26
Application of Arrhenius Equation
From k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T.
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy
Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?
Method: derive various versions of the same formula
k = A e – Ea / R T
A=ke
A/ k = e
Ea / R T
Ea / R T
Make sure you know how to
transform the formula into
these forms.
ln (A / k) = Ea / R T
15 Chemical Kinetics
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Application of Arrhenius Equation (continue)
The reaction:
2 NO2(g) -----> 2NO(g) + O2(g)
The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy
Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?
Use the formula derived earlier:
A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K –1*300 K))
= 2.13e9 s-1
k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K)
= 1.23e-12 s-1
T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1
= 285 K
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28
The Effect of Temperature on Reaction Rates
Reaction rate = k [A}x[B]y[C]z
(Concentration effect at constant T)
k = A exp ( – Ea / RT)
(Temperature effect)
Use graphic method to discuss the
variation of k vs. T
variation of k vs 1 / T
variation of ln(k) vs T
variation of ln(k) vs 1 / T
See a potential multiple choice question in an exam?
15 Chemical Kinetics
29
Elementary Reactions and Mechanism
Elementary reactions are steps of molecular events showing how
reactions proceed. This type of description is a mechanism.
The mechanism for the reaction between CO and NO2 is proposed to be
Step 1 NO2 + NO2  NO3 + NO
(an elementary reaction)
Step 2 NO3 + CO  NO2 + CO2
(an elementary reaction)
Add these two equations led to the overall reaction
NO2 + CO = NO + CO2 (overall reaction)
A mechanism is a proposal to explain the rate law, and it has to satisfy
the rate law. A satisfactory explanation is not a proof.
Explain terms in red
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Molecularity of Elementary Reactions
The total order of rate law in an elementary reaction is molecularity.
The rate law of elementary reaction is derived from the equation. The
order is the number of reacting molecules because they must collide to
react.
A molecule decomposes by itself is a unimolecular reaction (step);
two molecules collide and react is a bimolecular reaction (step); &
three molecules collide and react is a termolecular reaction (step).
O3  O2 + O
rate = k [O3]
NO2 + NO2  NO3 + NO
rate = k [NO2]2
Br + Br + Ar  Br2 + Ar*
rate = k [Br]2[Ar]
Caution: Derive rate laws this way only for elementary reactions.
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Molecularity of elementary reactions - Example
Some elementary reactions for the
reaction between CH4 and Cl2 are
Cl2  2 Cl
2 Cl  Cl2
2Cl + CH4  Cl2 + CH4*
Cl + CH4  HCl + CH3
Write down the rate laws
and describe them as
uni- bi- or ter-molecular
steps yourself, please.
CH3 + Cl  CH3Cl
CH3 + CH3  CH3-CH3
CH3Cl + Cl  HCl + CH2Cl
CH2Cl + Cl  CH2Cl2
* * * (more)
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Elementary Reactions are Molecular Events
N2O5  NO2 + NO3

NO + O2 + NO2  NO2 + NO3
Explain differences between elementary and over reaction equations
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33
Rate Laws and Mechanisms
A mechanism is a collection of elementary steps devise to explain the the
reaction in view of the observed rate law. You need the skill to derive a
rate law from a mechanism, but proposing a mechanism is task after you
have learned more chemistry
For the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), the rate law is,
rate = k [NO2] [F2] .
Can the elementary reaction be the same as the overall reaction?
If they were the same the rate law would have been
rate = k [NO2]2 [F2],
Therefore, they the overall reaction is not an elementary reaction. Its
mechanism is proposed next.
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Rate-determining Step in a Mechanism
The rate determining step is the slowest elementary step in a mechanism,
and the rate law for this step is the rate law for the overall reaction.
The (determined) rate law is,
rate = k [NO2] [F2],
for the reaction,
2 NO2 (g) + F2 (g)  2 NO2F (g),
and a two-step mechanism is proposed:
i NO2 (g) + F2 (g)  NO2F (g) + F (g)
ii NO2 (g) + F (g)  NO2F (g)
Which is the rate determining step?
Answer:
The rate for step i is rate = k [NO2] [F2], which is the rate law, this
suggests that step i is the rate-determining or the s-l-o-w step.
Explain rate determining step in a mechanism
and use it to derive the rate law.
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35
Deriving a Rate Law From a Mechanism - 0
The decomposition of H2O2 in the presence of I– follow this
mechanism,
i
ii
H2O2 + I–  k1 H2O + IO–
H2O2 + IO–  k2 H2O + O2 + I–
What is the rate law?
slow
fast
Energy
Eai
Eaii
reaction
Solve the problem
15 Chemical Kinetics
36
Deriving a rate law from a mechanism - 1
The decomposition of H2O2 in the presence of I– follow this mechanism,
i
ii
H2O2 + I–  k1 H2O + IO–
slow
H2O2 + IO–  k2 H2O + O2 + I–
fast
What is the rate law?
Solution
The slow step determines the rate, and the rate law is:
rate = k1 [H2O2] [I –]
Since both [H2O2] and [I –] are measurable in the system, this is the
rate law.
15 Chemical Kinetics
37
Deriving a rate law from a mechanism - 2
Derive the rate law for the reaction,
from the proposed mechanism:
i
Br2  2 Br
ii
H2 + Br  k2 HBr + H
iii
H + Br  k3 HBr
H2 + Br2 = 2 HBr,
fast equilibrium (k1, k-1)
slow explain
fast
Solution:
The fast equilibrium condition simply says that
k1 [Br2] = k-1 [Br]2
and
[Br] = (k1/k-1 [Br2])½
The slow step determines the rate law,
rate = k2 [H2] [Br]
Br is an intermediate
= k2 [H2] (k1/k-1 [Br2])½
= k [H2] [Br2] ½;
k = k2 (k1/k-1)½ M-½ s -1
total order 1.5
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38
Deriving a rate law from a mechanism - 3
The decomposition of N2O5 follows the mechanism:
1
N2O5  NO2 + NO3
fast equilibrium
2
NO2 + NO3 —k2 NO + O2 + NO2
slow
3
NO3 + NO —k3 NO2 + NO2
fast
Derive the rate law.
Solution:
The slow step determines the rate,
rate = k2 [NO2] [NO3]
From 1, we have
[NO2] [NO3]
—————— = K
[N2O5]
Thus, rate = K k2 [N2O5]
15 Chemical Kinetics
NO2 & NO3 are intermediate
K, equilibrium constant
K differ from k
39
Deriving rate laws from mechanisms
The steady-state approximation
is a general method for deriving
rate laws when the relative
speed cannot be identified.
It is based on the assumption
that the concentration of the
intermediate is constant.
– steady-state approximation
[Intermediate]
Rate of producing the
intermediate, Rprod, is
the same as its rate of
consumption, Rcons.
Rprod > Rcons
Be able to apply the steadystate approximation to derive
rate laws
Rprod = Rcons
Rprod < Rcons
time
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40
Let’s assume the mechanism for the reaction.
Steady-state
H2 + I2  2 HI
approximation as follows.
Step (1)
I2 —k1 2 I
Step (1)
2 I —k-1 I2
Step (2)
H2 + 2 I —k2 2 HI
Derive the rate law.
Derivation:
rate = k2 [H2] [I] 2 (‘cause this step gives products)
but I is an intermediate, this is not a rate law yet.
Since k1 [I2]
(= rate of producing I) Steady state
= k-1 [I]2 + k2 [H2] [I]2
(= rate of consuming I)
Thus,
k1 [I2]
[I]2 = ——————
k-1 + k2 [H2]

rate = k1 k2 [H2] [I2] / {k-1 + k2 [H2] }
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41
2
Steady-state approximation - 3
From the previous result:
k1 k2 [H2] [I2]
rate = ———————
{k-1 + k2 [H2] }
Discussion:
(i) If k-1 << k2 [H2] then {k-1 + k2 [H2]} = k2 [H2] ,
then rate = k1 k2 [H2] [I2] / {k2 [H2] } = k1 [I2] (pseudo 1st order wrt I2)
using large concentration of H2 or step 2 is fast (will meet this condition).
(ii) If step (2) is slow, then k2 << k1,
and if [H2] is not large, we have {k-1 + k2 [H2]} = k-1
and rate = k1 k2 [H2] [I2] / k1 = k2 [H2] [I2]
15 Chemical Kinetics
42
Steady-state approximation - 4
In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,
S2O82- + SO32- + 2 OH-  3 SO42- + H2O.
The following mechanism has been proposed:
i
S2O82- + SO32- —k1 S2O72- + SO42ii
S2O72- + H2O —k2 2 SO42- + 2 H+
iii
H+ + OH- —k3 H2O (fast equilibrium to be discussed)
Steady-state approximation follows these steps:
What is or are the intermediates I?
Use which step to give the rate law that may involve [I]?
Express the rates of producing and consuming intermediate(s)
Express [I] of intermediate(s) in terms of [Reactants]
Derive the rate law in terms of [Reactants]
Discuss
See page
607Kinetics
PHH Text
15 Chemical
43
Catalysis
Energy
A catalyst is a substance that changes the rate
Uncatalyzed rxn
of a reaction by lowing the activation energy, Ea.
It participates a reaction in forming an
intermediate, but is regenerated.
Enzymes are marvelously selective catalysts.
A catalyzed reaction,
NO (catalyst)
2 SO2 (g) + O2 — 2 SO3 (g)
via the mechanism
i
2 NO + O2  2 NO2 (3rd order)
ii
NO2 + SO2  SO3 + NO
15 Chemical Kinetics
Catalyzed rxn
rxn
44
Catalyzed decomposition of ozone
R.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds.
The CFC decomposes in the atmosphere:
CFCl3  CFCl2 + Cl
CF2Cl3  CF2Cl + Cl.
The Cl catalyzes the reaction via the mechanism:
i
O3 + h v  O + O 2,
ii
ClO + O  Cl + O2
iii
O + O3  O2 + O2.
The net result or reaction is
2 O3  3 O 2
Scientists sound the alarm, and the CFC is banned now.
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45
Homogenous vs. heterogeneous catalysts
A catalyst in the same phase (gases and solutions) as the reactants is a
homogeneous catalyst. It effective, but recovery is difficult.
When the catalyst is in a different phase than reactants (and products),
the process involve heterogeneous catalysis. Chemisorption, absorption,
and adsorption cause reactions to take place via different pathways.
Platinum is often used to catalyze hydrogenation
Catalytic converters reduce CO and NO emission.
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46
Heterogeneous catalysts
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Catalyzed reactions:
15 Chemical Kinetics
CO + O2 CO2
2 NO  N2 + O2
47
15 Chemical Kinetics
48
Enzymes – selective catalysts
Enzymes are a long protein molecules that fold into balls. They often
have a metal coordinated to the O and N sites.
Molecules catalyzed by enzymes are called substrates. They are
held by various sites (together called the active site) of the enzyme
molecules and just before and during the reaction. After having
reacted, the products P1 & P2 are released.
Enzyme + Substrate ES (activated complex)
ES  P1 + P2 + E
Enzymes are biological catalysts for biological systems.
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15 Chemical Kinetics
X-ray 3-D structure of
fumarate reductase. It
reduces fumerate, an
important role in the
metabolism of
anaerobic bacteria,
50
from Max Planck Inst.
Chemical Kinetics - Summary
Explain how the various factors affect reaction rates.
Define reaction rates, average rates, initial rates and rate constants.
Evaluate rate law from experiments
Properly apply 1st and 2nd differential rate laws and integrated rate laws.
Interpret elementary reactions and mechanisms. Derive rate laws from
a given mechanism. Apply the steady-state method to derive the rate
law of a given mechanism, and discuss the results.
Explain the action of catalysts in terms of chemistry and in terms of
energy of activation.
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