Download An elementary step is a reaction that proceeds as written

An elementary step is a reaction that
proceeds as written
This important because we know the rate law for an
elementary process.
RECALL:
REACTION MECHANISM: The process or series of
elementary steps by which a reaction occurs.
MJ Bojan
Chem 112
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If a step is elementary, then the rate of that reaction is
proportional to the concentration of the reactants.
Examples:
Rate Law
Unimolecular
Bimolecular
termolecular
PROBLEM: I can’t tell if a reaction is an elementary
process unless I do an experiment.
Experiments can be used to support a proposed
reaction mechanism or be proof that a
proposed mechanism is incorrect.
MJ Bojan
Chem 112
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Goal is to use kinetics to find the
Mechanism of a Reaction
What is the Mechanism for the following reaction?
NO2 +CO è NO + CO2
What would the rate law be IF it was an elementary
process?
"# [NO 2 ]
rate =
=
#t
What is the Experimental Result?
!
MJ Bojan
Chem 112
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Example: NO2 +CO è NO + CO2
Experimental shows:
Propose a mechanism:
NO2 + NO2 è NO3 + NO
1
NO3 + CO è NO2 + CO2
2
NO2 + CO è NO + CO2
Find the Rate Law for each step in a multi-step mechanism
!" [ NO2 ]
2
rate!1 =
= k [ NO2 ]
"t
!" [CO ]
rate!!2!=
= k [ NO3 ] [CO ]
"t
Then compare these with the experimental result.
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MJ Bojan
Chem 112
How do you know which step is the
Rate Determining Step
MJ Bojan
Chem 112
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Some Key points related to Mechanisms
•  Rate determining step = slow step
transition states
Energy
•  Elementary steps in a mechanism must
add up to give the balanced equation for
the overall process.
•  NO3 is produced in step 1 and
consumed in step 2
Intermediate: a stable molecule
Note:
intermediate
reactant
product
Reaction coordinate
•  Intermediates do not (should not) appear in the rate law.
(Concentration dependence of intermediates cannot be measured.)
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MJ Bojan
Chem 112
To find mechanisms
1.  Find the experimental rate law
2.  Postulate elementary steps
3.  Find the rate law predicted by the
mechanism and compare to experiment.
No rate can be written in terms of intermediates
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MJ Bojan
Chem 112
Mechanism Example Problem
Cl2 + CHCl3 è HCl + CCl4
Rate = kobs [Cl2]1/2 [CHCl3]
Postulate the following mechanism:
is it consistent with the experimental rate law?
Cl2
2Cl
fast
Cl + CHCl3 è HCl + CCl3
Cl + CCl3è CCl4
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slow
fast
MJ Bojan
Chem 112
CATALYSIS
Catalyst is a substance that speeds ups a reaction
without undergoing permanent change
How:
Thermodynamic state functions
(ΔE, Δ H, Δ G, Δ S…) are unaffected by catalysis
(Changes are path-independent)
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MJ Bojan
Chem 112
Catalysis
Homogeneous catalysis: when catalyst is in the same
phase as the reactant.
Heterogeneous catalysis: when catalyst is in a different
phase from the reactants
•  usually a solid catalyst and gas or solution reactants
•  reaction happens on the surface of the catalyst
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MJ Bojan
Chem 112
Demonstrate catalysis by studying the
decomposition of hydrogen peroxide.
Example:
2H2O2 (aq) è 2H2O () + O2(g)
Use 3 different catalysts:
Br− (I−)
MnO2
catalase
Homogeneous catalysis: (I−, catalase)
Heterogeneous catalysis: MnO2
Enzyme catalysis (catalase)
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MJ Bojan
Chem 112
Catalyze the decomposition reaction via
HOMOGENOUS CATALYSIS
Energy profiles for catalyzed and uncatalyzed
H2O2 decomposition
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MJ Bojan
Chem 112
Catalyze the decomposition reaction via
HETEROGENOUS CATALYSIS
EXAMPLE: H2 + 1/2 O2 è H2O
reaction requires breaking
strong H−H and O=O bonds
é
é
435 kJ
498 kJ
negligible rate without catalyst
solid
H-H
O=O
Pt surface
H
diffusion
on surface
H
H
H
O
=
O
=
H-H
=
adsorbed atoms
H H
O O
gas molecules
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MJ Bojan
Chem 112
CATALYTIC CONVERTER
Example of heterogeneous catalysis
O2
1) CO è CO2(g) + H2O
Hydrocarbons
2) NO, NO2
è N2(g)
Catalysts: CuO, Cr2O3, Pt, Rh
Usually, the stronger the bonds in reactants, the more we
need a catalyst.
e.g. 3H2 + N2 è 2NH3
ΔG° = −33 kJ/mol
N−N triple bond (D = 946 kJ)
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MJ Bojan
at 298 K
Chem 112
HOW DOES LOWERING Ea AFFECT RATES?
EXAMPLE.
H2O2 è H2O + 1/2 O2
hydrogen peroxide is toxic
Uncatalyzed reaction has Ea = 72 kJ
Catalase (enzyme in liver) lowers Ea to 28 kJ
What is the ratio of kcat/kuncat at 37°C? (body temperature)
Ea,cat
!
RT
kcat
Acat e
!=!
Ea,uncat
!
kuncat
Auncat e RT
Assume Acat = Auncat
(Q: is this a good assumption?)
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MJ Bojan
Chem 112
CATALYZED VS UNCATALYZED
kcat
!=!e
kuncat
!
(Ea,cat !Ea,uncat )
RT
kcat
!(28 ! 72)!kJ / mol
ln
!=!
!=!17.1
kuncat (0.0083!kJ / mol ! K )(310K )
kcat
"!!
!=!3#!10 7
kuncat
speeds up by a factor of
30 million!
Peptidase enzymes – break up proteins into amino acids
(in your stomach) have a similar effect on Ea
Without these it would take ~ 300 years to digest a steak
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MJ Bojan
Chem 112