The above is a bimolecular elementary reaction.
A unimolecular elementary reaction might be
HO2 → H + O2
HO2 just dissociates without any other influence.
Rate Laws for Elementary Reactions:
1) A → Fragments, depends only on A
dC A
∴−
= kCA
dt
[Like radioactive decay]
2) A+A → Products, depends only on A, A collision
dC A
2
=−
= kCA
dt
A + B → Products.
−dC A −dCB
=
= kCA CB .
dt
dt
Elementary reaction is one place where stoichiometry and
rate are related. However never know when you have an
elementary reaction. Must guess and then verify with experiment.
Elementary reactions are hypothetical constructs!
Unimolecular Decompositions
An example of
Mechanisms, Steady State Approximation,
and Elementary Reactions
A → Fragments
Observed Experimentally to
be first order in pyrazine.
pyrazine
dA
−
= k[A] or C - B → C + B with
dt
d[C − B]
−
= k[C − B]
dt
Lindemann [Lord Cherwell] suggested the following mechanism:
Collisions in general produce molecules with higher than average
energy. Consider a box full of A and let those molecules which
have an energy E* greater than or equal to that necessary for
decomposition be A*.
Assume, however, that after A* is produced by a collision it hangs
around for some time before decomposing. This time lag between
activation and reaction may be thought of as the time necessary to
transfer energy among the internal (vibrational) coordinates.
If A* exists for a reasonable time it could suffer a collision and drop
down to a lower energy where it cannot decompose.
Collision between
k1
A(E1) + A(E2)
A*(E*) + A(E3)
A(E1) and A(E2)
creates “activated”
Collision
A*(E*)
E1 , E2 , E3 < Emin; E*≥ Emin = energy necessary for decomposition
k −1

→
Competition between
reaction of A*(E*) to

Collional cooling
form products and
k 2
step
collisional cooling of

A*(E*) to
↓
Reaction step
produce unreactive
Products
A(E5) and A(E6)
Again E4 , E5 , E6 < Emin (k1, k-1, and k2 are kinetic rate constants)
A*(E*)
+ A(E4)
A(E5) + A(E6)
k2 is decomposition step assumed irreversible.
Mechanism Elementary Steps:
k1
Step 1
A + A 
→ A∗ + A
k
Step 2
A 
→ P
Step 3
−1
A ∗ + A 

→A + A
∗
k2
dP
= k2 [A∗ ]
dt
Step 1
Step 2
Step 3
d[A∗ ]
∗
∗
= k1[A]2 − k −1 [A ][A] − k2 [A ]
dt
Don’t know what [A*] is, however the number of [A*] must be small
or the reaction would go to completion very quickly. A* is a
“bottleneck” for the reaction since product is only formed via A*.
Therefore we assume a steady state for [A*] →
important “trick” or approximation:
d[A*]/dt = 0 = k1[A]2-k-1[A*][A]-k2[A*]
This is a most
The Steady State
Approximation
(Assumes are making and destroying A* at the same rate.)
Solve for [A*] →
[A*] = k1[A]2 / ( k2+k-1[A] ) } Key Point:
Steady state approach allows us to solve for concentration
of unknown species [A*] in terms of known [A] concentration.
2
Fundamental result of
k
dP
k
A
[
]
∗
2 1
Rate =
= k2 [A ] =
Lindemann “Unimolecular
dt
k2 + k−1 [A]
Reaction Mechanism”
Note, multistep mechanism leads to complex rate expression!
Two Limiting Cases
I) k-1 [A] << k2
2
[A]
dP
k
k
Rate =
= k2 [A∗ ] = 2 1
dt
k2 + k−1 [A]
This says the rate of decomposition of A* is much faster
than the rate of deactivation.
Using k2>>k-1[A] and the expression for dP/dt:
Take k-1[A] to be 0
compared to k2
Rate = dP/dt = k2[A*] = k2k1[A]2/(k2+k-1[A])
Thus, dP/dt ≅ k1[A]2 }2nd order in [A]
At this point, it looks like Mr. Lindemann will have to hand
in his Theorists’ Club ID card since his scheme seems to
predict a second order kinetic dependence on [A]2!
“Physical Interpretation” of this particular limit:
Mechanism Elementary Steps:
k1
Step 1
d[A*]/dt = k1[A]2
A + A 
→ A∗ + A
k
Step 2
Slow enough to be ignored!
A 
→ P
Step 3
So fast that every A* formed
Reacts immediately!
−1
A ∗ + A 

→A + A
∗
k2
Reaction rate here is just the rate at which [A*] is formed since
every A* formed falls apart to product P immediately. The rate of
formation of A* is obtained from the first step: d[A*]/dt = k1[A]2
So the reaction becomes a simple binary collison model, second
order process in this limit.
(Case II) k-1 [A] >> k2
dP
k 2 k1 [A]2
∗
Rate =
= k2 [A ] =
dt
k2 + k−1 [A]
This says there is an appreciable time lag between activation and
reaction. Thus, a large amount of deactivation occurs.
Take k2 to be 0
2
Rate = dP/dt = k2[A*] = k2k1[A] /(k2+k-1[A]) compared to k [A]
-1
Note that k-1 [A] is pressure dependent. Gets larger as pressure
increases {PA=(nA/V)RT = [A]RT}. Thus, k-1 [A] >> k2 is a good
approxmation at high pressures.

dP k2 k1
≅
[ A] 1st order in A (apparent)
k −1
dt

∴ At high pressures expect to see this behavior.
[More careful investigation of “unimolecular” decomp. showed 2nd
order kinetics at low pressure.]
“Physical Interpretation” of this particular limit:
Mechanism Elementary Steps:
k1
Step 1
A + A 
→ A∗ + A
2
k
[A*][A]
=
k
[A]
-1
1
k −1
∗
Step
2
A + A 
→ A + A
∗
k2
A 
→ P
Step 3
So slow that steps 1 and 2
Can “reach equilibrium”.
Reaction rate here is dP/dt = k2 [A*] but the concentration of A*
is given by the “equilibrium” condition, k-1[A*][A] = k1[A]2. So,
solving for [A*] gives:
[A*] = k1[A]2/ k-1[A] = k1[A]/ k-1
Result is that [A*] scales linearly with [A] and rate, dP/dt = k2[A*]
also scales linearly with [A].
Catalysis
Catalysis provides an additional mechanism by which reactants
can be converted to products. The alternative mechanism has a
lower activation energy than the reaction in the absence of a
catalyst.
A
υo
υc
B
υo no catalyst
υc -- catalyst present
(v0 = -d[A]/dt with no catalyst)
(vc = -d[A]/dt with a catalyst)
Without a catalyst: Rate = v0
With Catalyst: Rate=v0+vc but vc>>v0
Potential
Energy here
is energy
stored in the
chemical
bonds
E
Potential Energy
∆E Not affected
by catalyst
a,f
Energy barrier
with catalyst
Energy barrier without catalyst
Ea,f
Ea,f and Ea,r
are lowered
by
catalyst
Ea,f
Ea,r
∆E
Ea,r
Reactants
Products
Reaction coordinate
Catalysts speed reactions by reducing the activation
energy barrier.
Generally a catalyst is defined as a substance which increases the
rate of a reaction without itself being changed at the end of the
reaction.
This is strictly speaking not a good definition because some things
catalyze themselves, but we will use this definition for now.
Catalyst supplies a reaction path which has a lower activation
energy than the reaction in the absence of a catalyst.
Catalysis by Enzymes
Enzymes may be loosely defined as catalysts for biological systems.
They increase the rate of reactions involving biologically important
systems.
Enzymes are remarkable as catalysts because they are usually
amazingly specific (work only a particular kind of reaction.)
They are also generally very efficient, achieving substantial
Rate increases at concentrations as low as 10-8 M!
Typical enzyme molecular weights are 104-106 gm/mole
(protein molecules)
Summary of Enzyme Characteristics
1) Proteins of large to moderate weight 104 - 106.
2) Extremely efficient (work at 10-8 M)
3) Very specific (work only on special types of reactions).
General Behavior of Enzyme Catalyzed Reactions
S (substrate)
E
→ Products
If the initial rate of the reaction is plotted versus the initial
concentration of substrate S for a constant enzyme concentration,
the following behavior is found:
Maximum initial rate
Initial
Rate
(Half maximum initial rate)
[S] concentration when Vi = VS / 2
Substrate Concentration
Vi = - d[S]/dt measured at t=0. (Initial slope of [S] vs t plot)