I. Rates and Rate Laws Definition of Reaction Rate: So far we have ...

I. Rates and Rate Laws
Definition of Reaction Rate:
So far we have talked about the thermodynamics of process and reactions, as well as their
equilibrium. We now want to consider the timescales with which chemical and physical
processes occur. The study of the rates of reactions is called kinetics.
The first thing we have to do is define a rate, or velocity, of reaction. This can be expressed
as the time derivative of the concentration for a particular component. In this formalism, the
velocity of the reaction is related to a change in concentration with time:
dc
v= = rate
dt
Note that for a particular reaction, the concentrations of each component are coupled, so we
can express the rate in terms of the time rate of change for any reactant or product. In other
words, if a reactant transforms into product, the concentrations at any point in the reaction
are related by their stoichiometric coefficients. As a result, for the reaction:
aA + bB → cD + dD
The rate can be expressed in terms of any component:
−1 dA [ ] −1
dB [ ] 1 dC [ ] 1 dD [ ]
rate = = = =
a dt b dt c dt d dt
Note that the negative derivative for the reactants corresponds to their disappearance and the
positive derivative for the products corresponds to their appearance.
Rate Laws:
We can also express the rate in terms of the concentrations of the stoichiometric reactants
and products (i.e. those that appear in the net reaction.) In this way, the reaction rate can be
thought of in terms of collisions – if there is a higher concentration of a particular
component, then it is more likely to react, but there is some proportionality that takes into
account physical parameters of the system. We will discuss this later. For now, let’s
represent the reaction rate in terms of a rate law.
Example – consider the reaction:
The rate law for this reaction is:
A+ B→ P
m n q
rate =
k[ A] [ B] [ P]
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Where k is the rate constant and m , n and q are reaction orders with respect to a
particular component. For example, if m = 2 and n = 1,
we say that the reaction is second
order in A and first order in B. Reaction orders are typically integers (i.e. 0, 1, 2, …), but
they could be fractions.
Since the rate of reaction has units of concentration/time (M/s), the units for k are:
1 1
k[
] x 1
s M −
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
where x is the total reaction order, or the sum of the individual reaction orders.
II. Temperature Dependence of Rate Constants
Preliminaries – Arrhenius equation:
We would now like to consider the temperature dependence of the rate constants in our
reactions. Qualitatively, we expect that the reaction will occur with a faster rate if we
increase temperature, i.e.: there will be more collisions, thus a faster reaction. It has been
found empirically that the rate constants for reactions follow the Arrhenius Equation:
Ea/ RT
k Ae −
=
where A is called the pre-exponential factor (weakly T-dependent), Ea is called the
activation energy and R = 8.314 J/mol K is the universal gas constant.
The Arrhenius equation can be rearranged into:
Ea
ln k = ln A− RT
The rate constants at two different temperatures are related by the expression:
1 1
ln
E k ⎛ ⎞ ⎛ ⎞
=− −
2
a
⎜ ⎟ ⎜ ⎟
⎝ k1 ⎠ R ⎝T2 T1⎠
and if you have the rate constants at many temperatures, you can plot ln k vs. 1/T to obtain
a linear relationship where the slope is equal to − E / R and the intercept is equal to ln A .
The activation energy is typically on the order of 50 – 200 kJ/mol (weaker than a chemical
bond) and the pre-exponential factor is on the order of 10 10 – 10 14 s -1 , and is related to the
collision cross section. We will discuss this later.
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a

Energetics of activation – changing reaction rate with temperature:
We first want to understand what the activation energy is and how we can affect the rate of a
reaction. We can express energetic changes throughout a reaction on a potential energy (or a
reaction coordinate) diagram
The reaction rate is dependent on the height of the activation barrier. By increasing the
temperature of the reaction, the thermal energy of the system increases and it becomes more
probable that a particular collision can overcome the activation barrier, resulting in a faster
reaction, consistent with empirical observations. Be careful, though, because the equilibrium
conditions depend on the thermodynamic changes, independent of the activation barrier!
Catalysis – changing reaction rate by changing Ea:
Having defined the activation energy, we can consider ways to manipulate the reaction rate
by decreasing the activation barrier. One common way to speed up a reaction is to add a
catalyst. Energetically, this serves to decrease the activation barrier, making it more
probable that a particular collision can overcome the lowered barrier. Mechanistically, a
catalyst is a substance that is consumed in an early step in the mechanism and produced in a
later step. A catalyst may appear in the rate law even though it is not necessarily a
stoichiometric product or reactant, and it has been observed that reaction rates do typically
depend on the catalyst concentration.
One example is the Michaelis-Menten mechanism of enzyme catalysis. The mechanism is
as follows:
E + S←⎯⎯ ⎯⎯→ ES
k1
k−1
k2
ES ⎯⎯→ E+ P
Notice in this mechanism that the net reaction is S → P,
where the enzyme E is the catalyst
(consumed initially, then formed) and ES is the intermediate. Physically, ES is a complex
between enzyme and substrate.
The rate law can be determined using the SSA (since we have no information about relative
rates):
kk 1 2 rate = [ E][ S]
k + k
−1
2
This rate law is first order in catalyst concentration. The total enzyme (catalyst)
concentration occurs either as bound or unbound, and the total amount is conserved, so
[ E] = [ E] +
[ ES]
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0

This relation can be used in the SSA step to obtain a more useful form:
where V k2[ E]
0
= and ( )
−1
2 1
VS [ ]
rate =
K + [ S]
K = k + k / k is the Michaelis Constant.
m
Accounting for the pre-exponential factor – collision theory:
We now want to turn our attention to the physical meaning of the pre-exponential factor A .
This can be described by the kinetic theory for collisions. Consider the simple bimolecular
reaction in the gas phase:
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m
k
A+ B⎯⎯→ P
This reaction rate constant is determined by two parameters: A and Ea. In order for a
reaction to occur, A and B molecules have to collide with enough energy to overcome the
activation barrier. We expect that the number of collisions per unit volume per unit time
1/2
⎛8kT B ⎞
(collision density) is related to the RMS velocity of the gas particles urms
= ⎜ ⎟ where
⎝ π m ⎠
−23
kB= 1.38× 10 J / K is Boltzmann’s constant and m is the mass of the particle, the number
of particles N i and the collision cross-section σ . The collision cross basically says that if
two particles are within some critical area, there will be a collision.
If the two particles are within ( ) 2
σ π R R
= A + B , there will be a collision.
Now, the collision density Z AB between two unlike particles is:
1/2
⎛8kT B ⎞ 2
AB σ
av
Z = ⎜ ⎟ N [ A][ B]
⎝ πμ ⎠
where N av is Avogadro’s number and μ is the reduced mass and
1 1 1
= + .
μ m m
1 2
Now, the rate of change in the molar concentration of [A] is equal to the collision density
multiplied by the fraction of collisions that occur with a kinetic energy in excess of some
threshold value Ea:
dA [ ] ZAB
− = ×
f
dt N
av

where
Now,
f e −
Ea/ RT
= is given by the Boltzmann distribution of particles with energy Ea.
dA [ ] ⎛8kT B ⎞
− = σ ⎜ ⎟
dt ⎝ πμ ⎠
1/2
−Ea/
RT
Nave A B
[ ][ ]
Comparing this with the rate law rate = k[ A][ B]
, we find that:
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1/2
⎛8kT B ⎞
a
k = σ ⎜ ⎟ Nave ⎝ πμ ⎠
−E
/ RT
Comparing this with the Arrhenius form of the rate constant we find that:
1/2
⎛8kT B ⎞
A= σ ⎜ ⎟ N
⎝ πμ ⎠
as long as the exponential temperature dependence dominates the weak square root
dependence of the pre-exponential factor.
This form is slightly different than that used in LMS. They express the rate in terms of the
numbers of particles, while this representation uses molar concentrations. They also define
dAB = ( RA+ RB) = σ / π , and group the π term into the contribution from u rms .
Kinetic isotope effect:
From collision theory, we can see that the rate constant depends on a quantity called the
reduced mass μ . If you consider the relative rate constants between different isotopes, you
expect qualitatively that there should be a difference. For example, if you are running a
reaction where you break a C-H bond, you may expect a different rate if you broke a C-D
bond instead, because of the differences in reduced masses and the relative location of the
zero-point energies. This difference is rate constants results in the kinetic isotope effect,
which states that different isotopes react with different rates. This difference is most
pronounced when replacing H with D (or T), and is quantified by a quantity called the
isotopic ratio kH/ k D . Table 9.2 in LMS lists some values for isotopic ratios for H as well as
other atoms. For heavier elements, the kinetic isotope effect is much less pronounced.
Thermodynamic description of kinetics? Eyring transition state theory:
Another treatment used to describe the temperature dependence of reaction rates is called
activated complex theory, or transition state theory. Consider the simple bimolecular
reaction:
av

k
A+ B⎯⎯→ P
This reaction can be represented as a two-step process:
†
K †
† k
A+ B AB ⎯⎯→ P
†
where the species AB is called an activated complex, or transition state. Note that the
transition state is not the same thing as an intermediate!
Our goal now is to try to relate kinetic parameters to thermodynamic ones.
The first step in the process forms a fast equilibrium, so
considered to be irreversible.
Solving the rate law:
† †
rate = k [ AB ]
†
† [ AB ]
= and the second step is
[ A][ B]
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K
† †
rate = k K [ A][ B] = k [ A][ B]
† †
Now we only consider kobs = k K . We know that the Gibbs energy between the reactants
†
and the transition state is related to K :
Eyring theory states that the rate constant
obs
Δ G =− RTln K
† †
Planck’s constant. Now, the net rate constant is:
But since Δ G =ΔH −TΔ S,
If we compare this to the Arrhenius equation:
†
B
k is equal to
kT
h
†
B −ΔG
/ RT
kobs = e
kT
, where
h
−34
h 6.626 10 J s
= × i is
† † kT B
−Δ ⎡ H −TΔS ⎤/
RT
⎣ ⎦
kobs = e
h
kT † †
B ΔS / R −ΔH
/ RT
kobs = ⎡e ⎤⎡e ⎤
h ⎣ ⎦⎣ ⎦ [1]
Arrhenius: A plot of ln k vs. 1/T yields a slope of − E / R
†
TS Theory: A plot of ln k vs. 1/T yields a slope of −( Δ +
)
a
H RT / R

By comparison:
E = Δ H + RT ΔH
a
† †
since Ea~ 100 kJ / mol and RT ~2.5 kJ / mol . Plugging this into equation [1] and
comparing with the Arrhenius form, we find that:
†
⎡ ΔS
/ R⎤
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kT B
A= e
h ⎣ ⎦
Thus from information about the reaction rates at different temperatures, we can apply TS
theory to obtain information about the energetics of the transition state in the reaction!
III. Integrated Rate Laws
Rate Laws:
We can also express the rate in terms of the concentrations of the stoichiometric reactants
and products (i.e. those that appear in the net reaction.) In this way, the reaction rate can be
thought of in terms of collisions – if there is a higher concentration of a particular
component, then it is more likely to react, but there is some proportionality that takes into
account physical parameters of the system. We will discuss this later. For now, let’s
represent the reaction rate in terms of a rate law.
Example – consider the reaction:
The rate law for this reaction is:
A+ B→ P
m n q
rate = k[ A] [ B] [ P]
Where k is the rate constant and m , n and q are reaction orders with respect to a
particular component. For example, if m = 2 and n = 1,
we say that the reaction is second
order in A and first order in B. Reaction orders are typically integers (i.e. 0, 1, 2, …), but
they could be fractions.
Since the rate of reaction has units of concentration/time (M/s), the units for k are:
1 1
k[
] x 1
s M −
⎛ ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
where x is the total reaction order, or the sum of the individual reaction orders.
Integrated Rate Laws:

Zero Order Reactions:
Lets say the reaction A → B has the rate law
rate k A
dA [ ]
− = rate = k
dt
0
= [ ] . Since
we can solve the differential equation for the concentration of A with time:
if we define [A] = [A]0 at time t = 0, then:
[ A] dA [ ] = − kdt
∫ dA [ ] =−∫
kdt
− [ A] =−k( t − t )
f i f i
[ A] = [ A] − kt
A similar procedure can be done to determine the appearance of B with respect to time.
First Order Reactions:
Lets say the reaction A→ B has the rate law
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0
rate k A
dA [ ]
− = kA [ ]
dt
1
= [ ] . Since
we can solve the differential equation for the concentration of A with time:
if we define [A] = [A]0 at time t = 0, then:
In this reaction, [A]0 + [B]0 = [A] +
dA [ ]
= − kdt
[ A ]
dA [ ]
kdt
[ A ]
=− ∫ ∫
[ A]
f
ln = −kt ( f − ti)
[ A ]
i
[ A] [ A] e −
=
0
kt

[B], so if [B] = 0 at time t = 0,
Second Order Reactions:
Type I:
kt ( )
[ B] [ A] 1 e −
= −
Lets say the reaction 2A→ B has the rate law
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rate k A
1 dA [ ]
− = kA [ ]
2 dt
2
= [ ] . Since
we can solve the differential equation for the concentration of A with time:
if we define [A] = [A]0 at time t = 0, then:
Type II:
dA [ ]
2kdt
2
[ A ]
=−
dA [ ]
2kdt
2
[ A ]
=−
∫ ∫
1 1
= + 2kt
[ A] [ A]
Lets say the reaction A + B→ C has the rate law rate = k[ A][ B]
. Since
dA [ ]
− = kA [ ][ B]
dt
if we let [ A] = [ A] 0 − x and [ B] = [ B] 0 − x,
then we can perform a change of variables:
and
dx
rate = = k[ A][ B] = k [ A] 0 −x[ B] 0 − x
dt
dx
[ A] x [ B] x =
( − )( −
)
0 0
0
2
( )( )
kdt

Solving using partial fractions for [ A] 0 ≠ [ B]
0
finally
If 0 0
( − )
( )
1 [ B] 0 [ A] 0 x
ln
[ A] −[ B] [ A] [ B] −x
0 0 0 0
1 [ B] [ A]
[ ] [ ] [ ] [ ]
0 ln
A 0 B 0 A 0 B =
−
[ A] = [ B]
, then the expression reduces to Type I.
n th Order Reactions:
= kt
Lets say the reaction nA → B has the rate law [ ] n
rate = k A . Then
IV. Mechanisms
Goals:
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kt
1 ⎛ 1 1 ⎞
⎜ ⎟
n−1 ⎝[ A] [ A]
⎠
− n−1 n−1 0
= nkt
We have previously discussed reactions that occur in one step. We now want to turn our
attention to more complex reaction mechanisms that occur over many steps. The set of steps
in the reaction is called the reaction mechanism. The reaction mechanism consists of a
number of elementary reactions.
The eventual goal in this process is to propose a mechanism where the rate law is consistent
with experimental data. It is useful to point out on the onset that the rate law does not prove
a particular mechanism, but it can disprove the mechanism. In other words, two different
mechanisms may result in the same rate law.
Parallel Reactions:
Consider the following reaction, where the reactant can form either product B or product C:
k1
A⎯⎯→ B
k2
A⎯⎯→C

We can express the rate of laws for each component:
dB [ ]
k [ A]
1
dt = , 2
dt
dC [ ]
−dA
[ ]
= k [ A]
and = k1[ A] + k2[ A] = ( k1+ k2) [ A]
dt
We can integrate the last expression to get the concentration profile of [A]:
Now:
dB [ ]
dt
[ A] = [ A] e
− ( k1+ k2) t
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dC [ ]
dt
− ( k1+ k2) t
− ( k1+ k2) t
= k1[ A] = k1[ A] 0e
and = k2[ A] = k2[ A] 0e
If we assume that at t = 0, [B]0 = [C]0 = 0, then we can integrate the above expressions to
obtain the concentration profile of [B] and [C]:
Note that for all times:
Equilibrium Reactions:
k [ A]
B = −e
1 0 [ ] 1
k1+ k2
− ( k1+ k2) t { }
[ B] k
=
[ C] k
Consider the equilibrium process between A and B:
k [ A]
k + k
2 0 − ( k1+ k2) t
and [ C] = { 1−e
}
1
2
A←⎯⎯ ⎯⎯→ B
k1
k−1
1 2
In this reaction, the reactant A forms B, but then B converts back to A. We denote the rate
constant for a reverse reaction with a negative subscript. Consider the rate law for [A]:
−dA
[ ]
rate = = k1[ A] − k−1[ B]
dt
In this expression, the reaction k1 results in a disappearance of [A] (negative derivative),
while the reaction k-1 results in formation of [A] (positive derivative).
−dA
[ ]
When the reaction reaches equilibrium, we know that = 0 , i.e.: there is no net
dt
formation or consumption of [A]. Relating this to the rate law:

But we also know that [ B] /[ A] = K , so:
eq eq eq
[ B] eq k1
=
[ A] k− K
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eq
eq
k
=
k
1
forward
reverse
I smell a relationship with thermodynamics… Don’t get too excited.
Consecutive Reactions:
Now consider a reaction that occurs in two consecutive steps:
k1 k2
A⎯⎯→B⎯⎯→ C
The rate laws for each expression are as follows:
−dA
[ ]
dt
dB [ ]
k [ A] k [ B]
dt = − and dC [ ]
k2[ B]
dt =
= k1[ A]
, 1 2
In the above expressions, [A] disappears with rate constant k1, [B] is formed as k1, but
disappears (negative derivative) as k2, and [C] is formed as k2.
We can integrate these rate laws to get the respective concentration profiles (assuming [B]
and [C] are zero initially):
Now,
[ A] [ A] e −
=
0
kt 1
dB [ ]
−kt
1
= k1[ A] − k2[ B] = k1[ A] 0e − k2[ B]
dt
dB [ ]
−kt
1
+ k2[ B] = k1[ A] 0e
[1]
dt
To solve this equation, we first make an assumption as to the functional form of [B]:
[ B] yte ( )
−
kt 2
t = [2]
This basically says that [B] decays exponentially, but with a time-dependent amplitude. The
‘dummy’ function y(t) basically accounts for the buildup of [B] from reaction 1, but we have

to find what y(t) is. With [2] and the chain rule, we can take the time derivative of this form
for [B]:
dB [ ]
−kt 2 −kt⎛dy⎞ 2
=− yt () ( ke 2 ) + e ⎜ ⎟
dt ⎝ dt ⎠
But the first term is simply equal to –k2[B]t, so moving to the other side:
Equating [1] and [3]:
dB [ ]
−kt⎛dy⎞ 2
+ k2[ B] t = e ⎜ ⎟
dt ⎝ dt ⎠ [3]
⎛dy ⎞ −kt 2 −kt
1
⎜ ⎟e
= k1[ A] 0e
⎝ dt ⎠
⎛dy ⎞
⎜ ⎟=
k [ A] e
⎝ dt ⎠
( e ) = k [ A] e
− kt 1 + k2t ( k2−k1) t
1 0 1 0
k[ A]
y() t = e + C
k − k
1 0 ( k2−k1) t
Integrating:
2 1
k1[ A]
0
We know that at t = 0, y = 0, so: C =−
k − k
k [ A]
k − k
2 1
1 0 ( k2−k1) t
So: yt () = ( e −1)
kt 2
Finally, since [ B] yte ( )
−
= :
2 1
k [ A]
[ ] = −
−kt 1 −k2t
( )
1 0 B e e
k2 − k1
To get [C], we know that [A] + [B] + [C] = [A]0, so:
⎡ ke − ke ⎤
[ C] = [ A]
1−
−kt 1 −k2t
2 1
0 ⎢ ⎥
⎣ k2 − k1
⎦
This is the last differential equation we will solve for a while…
Let us consider two extremes in this 2-step mechanism:
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(1) First step is fast ( k1 k2)
[ ]
k[ A]
−kt 1 −k2t
( )
1 0 B = e −e
k2 − k1
−kt 1 −k2t
2 1
0 ⎢ ⎥
⎣ k2 − k1
⎦
k
[ B] = [ A] 0 − e = [ A] e
−k
1 → ( )
−kt 2 −kt
2
0 0
−kt
2
⎡ ke − ke ⎤
⎡ 0 − ke ⎤ 1
−kt
2
[ C] = [ A]
1−
→ [ C] = [ A] 0 ⎢1 − ⎥ = [ A] ⎡ 0 1−e
⎤
k ⎣ ⎦
⎣ − 1 ⎦
Consider this graphically:
(2) Second step is fast ( k2 k1)
:
k[ A]
[ ] = −
−kt 1 −k2t
( )
1 0 B e e
k2 − k1
−kt 1 −k2t
2 1
0 ⎢ ⎥
⎣ k2 − k1
⎦
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1
The reaction behaves as if the first
step did not exist – as if [B] is the
initial reactant (with a concentration
equal to [A]0) and the net reaction is
first order in [B]. Since the second
step dictates the kinetics of the entire
reaction, it is said to be the rate
determining (or rate limiting) step.
k[ A] k[ A]
B = e − = e
k k
→ [ ] ( 0)
1 0 −kt 1 1 0 −kt
1
2 2
−kt
1
⎡ ke − ke ⎤
⎡ ke ⎤ 2
−kt
1
[ C] = [ A]
1−
→ [ C] = [ A] 0 ⎢1 − ⎥ = [ A] ⎡ 0 1−e
⎤
k ⎣ ⎦
⎣ 2 ⎦
Consider this graphically:
This is equivalent to the
concentration profiles we would
obtain if the reaction were one-step
A → C . The concentration of [C]
behaves as if [B] were not there. In
this reaction, the first step is the rate
determining step.

Take home message: The overall rate of reaction is equivalent to the rate for the
rate limiting step!
We can use this to determine the rates for more complex reaction mechanisms.
Elementary Reactions
We have already had a taste of mechanisms. Now we can consider consecutive,
equilibrium and parallel reactions all at the same time. First, however, we have to
consider how the elementary reactions translate to rate laws. We do this by
determining the molecularity of the step.
The molecularity of a reaction is the number of molecules coming together to react.
For example, if we consider the isomerization of cyclopropane to propene:
We notice that a single molecule breaks itself into the new arrangement. A reaction
with a single molecule as a reactant is called a unimolecular reaction.
In a bimolecular reaction, a pair of molecules collide and exchange energy, such as
in the reaction of a hydrogen radical with gaseous bromine:
H i+ Br → HBr + Bri
2
The rate law for an elementary reaction simply takes the molecularity of the reaction
as the reaction order. For example, the rate law for the unimolecular reaction (first
reaction above) would be rate = k[ cyclopropane]
and the bimolecular reaction would
be rate = k[ H i ][ Br2
] . Note that we cannot do this for the net reaction, only for the
elementary reactions!
Steady State Approximation and Pre-equilibrium:
Consider the following 3-step mechanism for the net reaction A+ B+ D→ P:
A+ B←⎯⎯ ⎯⎯→ X
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k1
k−1
k2
X + D⎯⎯→ P
We do not necessarily know which step is rate limiting, nonetheless we can write the
rate law in terms of the formation of product P:

dP [ ]
rate = = k2[ X ][ D]
dt
Unfortunately, this rate law depends on the intermediate [X]. An intermediate is a
substance that is a product of an earlier step and a reactant in a later step.
Intermediates cannot appear in the overall rate law!
To eliminate the intermediate concentration from the rate law, we can use the steady
state approximation (SSA). If we consider the concentration of [X] as the reaction
proceeds, [X]0 = 0, but increases as the reaction begins. Once it is formed, it becomes
consumed in the second reaction. For most of the reaction, [X] is at a steady state
(constant) value. This means that the rate of formation for [X] is equal to zero! Now:
dX [ ]
≈ 0
[SSA]
dt
dX [ ]
= k1[ A][ B] −k−1[ X] − k2[ X][ D]
= 0
dt
k1[ A][ B] = k−1[ X] + k2[ X][ D]
k1[ A][ B]
[ X ] =
k + k [ D]
−1
2
We can use this steady state concentration of X to plug into our original rate law:
dP [ ] ⎛ k1[ A][ B] ⎞ kk 1 2[
A][ B][ D]
rate = = k2[ X ][ D] = k2⎜ ⎟[
D]
=
dt ⎝k−1+ k2[ D] ⎠ k−1+ k2[ D]
We now have a rate law in terms of stoichiometric products and reactants! This rate
law is true independent of the relative rates of each step
But… Consider the case where ( k−1 k2)
kk[ A][ B][ D]
1 2 rate = = k1 A B
k2[ D]
[ ][ ]
This is consistent with the rate law as if the mechanism were:
k1
A+ B⎯⎯→ P
So the rate law does not see the second step.
Now consider the case where ( k−1 k2)
:
D. Savin © 2009 PSC 480 Kinetics Supplement Page 16 of 33

kk[ A][ B][ D] kk
= = [ ][ ][ ] = [ ][ ][ ]
1 2 1 2
rate A B D kobs A B D
k−1 k−1
If we think about what is physically happening here, the reverse step of the first
reaction is large, so much so that the first reaction reaches equilibrium before the
second step begins!
Consider another way. We mentioned previously that
dP [ ]
rate = = k2[ X ][ D]
dt
If the first step reaches equilibrium rapidly (i.e.: before the second step begins), then:
K
1
[ X ]
= , so [ X ] K1[ A][ B]
[ A][ B]
= and 2 1 =
rate k K [ A][ B][ D]
This is consistent with the general expression above since K1 = k1/ k−1. This is called
the pre-equilibrium approximation. It is used whenever a mechanism involves a fast
initial equilibrium step.
Example: Determine the rate law for the following mechanism:
A←⎯⎯ ⎯⎯→ B+ C
(fast equilibrium)
k1
k−1
k2
k−2
k3
A+ C←⎯⎯ ⎯⎯→ E
(fast equilibrium)
E⎯⎯→ P
(slow)
From above, the net reaction is 2A → B+ P,
so C and E are intermediates. Since
step 3 is rate limiting:
dP [ ]
rate = = k3[ E]
dt
Using the pre-equilibrium approximation:
So:
[ E]
K
2
1 2 = and
1
KK[ A]
[ B]
[ B][ C]
[ E]
= and K2
=
[ A]
[ A][ C]
kKK[ A] [ A]
[ B] [ B]
2 2
3 1 2 rate = =
kobs
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Principles of Microscopic Reversibility and Detailed Balance:
For a series of equilibrium reactions, the net equilibrium constant is the product of the
elementary reaction equilibrium constants. This is also the product of the forward
rate constants divided by the product of the reverse rate constants. The Principle of
Microscopic Reversibility (Tolman) states that, “…in a system at equilibrium, any
molecular process and the reverse of that process occur, on average, at the same rate.”
A closely related principle states that at equilibrium, every molecular collision has an
exact counterpart in the reverse direction such that the rate in the forward direction is
balanced by the reverse reaction. This is called the Principle of Detailed Balance.
Note that both of these principles apply to equilibrium systems.
Checking for Consistency:
Armed with a procedure to determine rate laws for composite reactions, we can
consider how a proposed mechanism can be consistent (or not) with an
experimentally determined rate law.
+ −
Example: The reaction Cl2 + H2S→ S + 2H+ 2Cl
[ Cl2][ H2S] Has the experimentally determined rate law: rate k
[ H ]
+ =
Two mechanisms have been proposed. Which one is consistent with the observed
rate law?
Mechanism 1:
Mechanism 2:
←⎯⎯ ⎯⎯→ +
[fast equilibrium]
HS 2
k1
k−1
−
HS
+
H
k2
Cl ⎯⎯→ − +
2 Cl + Cl
k−2
+ − k3
+ −
←⎯⎯ [fast equilibrium]
Cl + HS ⎯⎯→+ H + Cl + S
[slow]
←⎯⎯ ⎯⎯→ +
[fast equilibrium]
HS + Cl ⎯⎯→ Cl + S + H
[slow]
HS 2
k1
k−1
−
HS
+
H
−
2
k2
2
− +
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Specific Mechanisms:
Lindemann – Hinshelwood mechanism:
Many gas phase reactions are unimolecular, for example the isomerization of
cyclopropane mentioned previously. The issue with first order reactions is that the
molecule somehow has to obtain enough energy (from collisions with other
molecules) for the reaction to occur (we will discuss activation energies soon!) But
collisions are bimolecular by nature, so one would think that they should not result in
a first order reaction. The mechanism for these type of reactions was provided by
Lindemann, then elaborated by Hinshelwood. In the mechanism, two reactant
molecules collide to form an activated molecule:
k1
A+ A⎯⎯→ A* + A
The activated molecule can suffer one of two fates. It may lose its energy through
another collision with an A molecule:
Or is can fragment into products:
k2
A* + A⎯⎯→ 2A
k3
A* ⎯⎯→ P
Of course, the first two reactions are reverse processes, so the net mechanism can be
written as:
A+ A←⎯⎯ ⎯⎯→ A* + A
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k1
k2
k3
A* ⎯⎯→ P
We do not have any information about the relative rates (at this point), so we cannot
use the pre-equilibrium approximation, but we can use the SSA:
so:
dP
rate = = k3[ A*]
dt
2
k1[ A]
[ A*]
=
k3 + k2[ A]
2
dP kk 1 3[
A]
rate = =
dt k + k [ A]
3 2
This mechanism certainly does not appear to be first order in [A], but we can consider
two extremes for the disintegration of [A*]:

Consider the case where the deactivation process is much faster than the
2
fragmentation ( k3 k2 [ A]
) . Then the rate becomes: rate = k1[ A]
But if the product 2 [ ] k A is much larger than k 3 :
kk 1 3
rate = kobs[ A]
where kobs
k
First order in [A]! It has been shown experimentally that many gas phase reactions
are second order in reactant at low concentration and first order in reactant at high
concentrations.
Chain reactions – the hydrogen-bromine reaction:
Consider the gas-phase reaction between hydrogen and bromine to form HBr:
H2 + Br2 → 2HBr
This reaction involves the formation of radicals, which are very reactive. One aspect
of radical reactions is that radicals beget radicals, meaning that the products of the
elementary reactions are typically radicals. These radicals continue to propagate until
a termination, typically by radical recombination. As a result, the mechanism for this
reaction can be expressed as a chain reaction:
D. Savin © 2009 PSC 480 Kinetics Supplement Page 20 of 33
2
k1
Br2⎯⎯→ 2Br
[initiation]
k2
Br+ H2⎯⎯→ HBr+ H
[propagation]
k3
H + Br2 ⎯⎯→ HBr + Br
[propagation]
k−2
H + HBr ⎯⎯→ Br + H 2
[retardation]
k−1
2Br⎯⎯→
Br
[termination]
The rate can be expressed as the formation of HBr, but because HBr is in many steps
in the mechanism:
2
dHBr [ ]
rate = = k2[ Br][ H 2] + k3[ H ][ Br2] − k−2[ H ][ HBr]
[1]
dt
We should go immediately to the SSA for the intermediates Br and H:

and:
dH [ ]
= k2[ Br][ H2] −k3[ H][ Br2] − k−2[ H][ HBr]
= 0
[2]
dt
dBr [ ]
2
= k1[ Br2] − k2[ Br][ H2] + k3[ H][ Br2] + k−2[ H][ HBr] − k−1[ Br]
= 0 [3]
dt
We have two equations and two unknowns. If we add [2] and [3]:
so:
k Br k Br
2
1[ 2] − −1[
] = 0
⎛ k ⎞
= ⎜ ⎟
⎝ ⎠
1 [ Br] [ Br2
]
k−1 We can plug this into [2] and rearrange to get [H]:
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1/2
1/2
⎛ k ⎞
1
2⎜ 2 ⎟ 2
k−1
k [ Br ] [ H ]
[ H ] =
⎝ ⎠
k [ Br ] + k [ HBr]
3 2 −2
Now, subtracting [2] from [1], and substituting the expression for [H], we obtain:
3/2
dHBr [ ]
⎛ k ⎞ 1 2 kk 3 2[ H2][ Br2]
rate = = [ H ][ Br2
] =⎜ ⎟
dt ⎝k−1⎠ k3[ Br2] + k−2[ HBr]
This can be rearranged to obtain:
where
2
1/2
3/2
kH [ 2][ Br2]
rate =
[ Br ] + k'[ HBr]
= 2
⎛ k1
2 ⎜
1
1/2
⎞
⎟
k k
⎝k−⎠ and
k
k ' =
k
Consistent with the experimentally observed rate law.
−2
3

V. Analysis of Kinetic Data
Differential Rate Method/Initial Rate Method:
As the name implies, this method is based on the idea that the rate can be expressed
as the time derivative of concentration for a particular component.
1 [ ]
[ ] n
dX
rate = = k X
x dt
So if you take the derivative of [X] with respect to time, you will obtain an
instantaneous rate. If you extrapolate this to the initial rate, you can compare the
initial rates of reaction for various initial concentrations of species that appear in the
rate law.
m n
Consider the generic reaction A+ B→ P given by the rate law kA [ ] [ B ] . If we run
the reaction under different sets of conditions, we will obtain different initial rates.
We can then compare these initial rates to obtain more information about the reaction
orders and the rate constant. For example, if we run two trials holding [B] constant
and varying [A], the initial rates are related by:
m n
rate1 k[ A] 1[ B] ⎛[ A]
⎞ 1
= =⎜ m n ⎟
rate2 k[ A] 2[ B] ⎝[ A]
2 ⎠
This procedure is then done for each component in the rate law.
Example: Consider the reaction of pyridine and methyl iodide:
C6H5N+ CH3I→ products
Initial rate data were taken at various concentrations of the reactants below:
Trial [ CHN] 6 5 (M) [ CH3I ] (M) Initial rate (M/s)
1
2
3
1.00 10 −
×
2.00 10 −
×
2.00 10 −
×
4
4
4
1.00 10 −
×
2.00 10 −
×
4.00 10 −
×
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4
4
4
m
7.5 10 −
×
3.0 10 −
×
6.0 10 −
×
From trials 2 and 3, if [A] remains constant and [B] doubles, the rate doubles, so the
reaction must be first order in [B]. From trials 1 and 2 if both initial concentrations
double, the rate quadruples, so the reaction must be first order in [A] as well.
Now that we have the reaction orders, we can calculate the rate constant:
7
6
6

ate
k =
[ A] [ B]
1 1
0 0
All three trials should yield the same rate constant, but to be sure we average over all
three trials. Finally:
−1 −1
( )
rate = 75 M s [ A][ B]
Some experimental error may be present in real data, so in the literature the rate
constants and reaction orders include error bars, i.e.: n = 1.02 ± 0.1 would correspond
to a reaction order of 1.
Integral Rate Method:
The integral rate method assumes you have data for [A] as a function of time. You
then guess the reaction order with respect to a particular reactant (or species in the
rate law) and plot the data in form consistent with the integrated rate law to yield a
linear fit.
1. Zero order reactions:
From the integrated rate law:
[ A] = [ A] − kt
A plot of [A] vs. time should yield a linear relationship where the slope of the line
is equal to − k and the intercept is equal to [ A ] 0 . If there is not a linear
relationship in the data, the reaction is not zero order with respect to [A].
2. First order reactions:
From the integrated rate law:
This can be rearranged to:
D. Savin © 2009 PSC 480 Kinetics Supplement Page 23 of 33
0
[ A] [ A] e −
=
0
0
kt
ln[ A] = ln[ A] −
kt

If the reaction is first order in A, A plot of ln[ A ] vs. time will yield a straight line
with a slope equal to k
ln[ A ] .
3. Second order reactions – Type I:
From the integrated rate law:
− and an intercept equal to 0
1 1
= + 2kt
[ A] [ A]
If the reaction is second order in [A], a plot of 1/[ A ] vs. time will yield a linear
plot with a slope of 2k and an intercept of 1/[ A ] 0 . Note that some treatments
group the ‘2’ (from the stoichiometry of the reaction) into the rate constant.
4. Second order reactions – Type II:
From the integrated rate law:
This can be rearranged into:
1 [ B] [ A]
ln
[ ] [ ] [ ] [ ]
0
A 0 B 0 A 0 B =
−
0
ln ⎜ ⎟ ln ⎜ ⎟ [ ] 0 [ ] 0
[ B] [ B]
0
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0
kt
⎛[ A]
⎞ ⎛[ A]
⎞
= + k( A − B ) t
⎝ ⎠ ⎝ ⎠
A plot of ln ( [ A]/[ B ] ) vs. time will yield a linear plot with a slope of
( [ A] − [ B] ) k and an intercept of ln ( [ A] /[ B ] ) .
0 0
5. n th order reactions:
0 0
From the integrated rate law (grouping stoichiometric coefficients together):
1 1
[ A] [ A]
− = kt
n−1 n−1 0
If the reaction is n th 1
order in [A], a plot of 1/[ ] n
A − vs. time will yield a linear plot
1
with a slope of k and an intercept of A − .
1/[ ] 0
n

Flooding/Isolation Method:
Consider the reaction A + B→ C where rate = k[ A][ B]
. The data for [A] vs. time
and [B] vs. time may be difficult to analyze. The integrated rate law is:
⎛[ A]
⎞ ⎛[ A]
⎞
= + k( A − B ) t
⎝ ⎠ ⎝ ⎠
0
ln ⎜ ⎟ ln ⎜ ⎟ [ ] 0 [ ] 0
[ B] [ B]
0
Now let us consider running this reaction with the condition that [ A] 0 [ B]
0 such
that the concentration of A remains nearly constant throughout the reaction. The
integrated rate law can be rearranged:
⎛ [ A] ⎞ ⎛ [ B]
⎞
ln⎜ ⎟− ln ⎜ ⎟=
k( [ A] 0 −[
B] 0)
t
[ A] [ B]
⎝ 0 ⎠ ⎝ 0 ⎠
But [ A]/[ A] 0 ≈ 1 and [ A] 0 −[ B] 0 ≈ [ A]
0.
Now:
And:
⎛ [ B]
⎞
ln ⎜ ⎟=−kA
[ ] 0t
[ B]
⎝ 0 ⎠
[ B] [ B] e −
=
k[ A] 0 t
Now the reaction looks first order in [B]. This technique of isolation results in a
pseudo-first order reaction. Physically, one component [A] is so large that its
concentration stays constant over the course of the entire reaction, so [ A] 0 ≈ [ A]
t .
This isolates the other component [B].
Now, a plot of ln[ B ] vs. time will yield a straight line with an intercept equal to
ln[ B ] 0 and a slope equal to − k ',
the pseudo-first order rate constant. The true second
order rate constant is simply k =
k'/[ A]
0
D. Savin © 2009 PSC 480 Kinetics Supplement Page 25 of 33
0

More complicated techniques – Kezdy-Swinbourne and Guggenheim Methods:
The analysis of kinetic data relies on the ability to measure the concentration of a
particular component in the reaction. Sometimes this is difficult to do. Consider the
thermal decomposition of chlorocyclohexane (CXCl):
CH Clg ( ) → CH ( g) + HClg ( )
6 11 6 10
In order to obtain the rate constant for this reaction, we would have to measure the
pressure of CXCl. This is extremely difficult to do since the reaction vessel contains
a mixture of gases. On the other hand, measuring the total pressure is relatively easy.
We need to develop a method to relate the rate constant of the reaction to the relative
total pressures at various points in time. In order to do this, we have to make
assumptions as to the reaction order with respect to the reactant. If we assume the
above reaction is first order in [CXCl], we can define some boundary conditions:
Let z t be the total pressure at some time t . Initially, zt= z0,
and at the end of the
reaction, zt= z∞. We now need a functional form of z t that satisfies these boundary
conditions while incorporating a single exponential decay characteristic of our first
kt
order reaction ( [ A] [ A] 0e
−
= ):
t
( )
z z z z e −
= + −
∞ 0 ∞
We do not necessarily know where z∞ is because of equilibrium conditions, unless
we do a good thermodynamic analysis, but we can relate the total pressure z t at some
time t with the total pressure zt+ τ at some later time t + τ .
Now, subtracting [1] from [2];
t
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kt
( 0 ) kt
( 0 ) − k( t+
τ )
( ) −kt −k
z z z z e −
= + − [1]
∞ ∞
z = z + z − z e
t+
τ ∞ ∞
z z z z e e τ
= + − [2]
t+
τ ∞ 0 ∞
t+ τ t ∞
−kt −k
( 0 ) ( 1 )
z z z z e e τ
− = − − [3]
⎡ ⎤
⎣ ⎦
k
and ln ( zt zt) kt ln ( z z0)( 1 e ) τ −
+ τ − =− + ∞ − −
Now we can plot ( z z )
a Guggenheim Plot.
ln t+ τ t − vs. time and expect a linear relationship. This is called

From this plot, the slope is equal to − k and the intercept (t = 0) is equal to
ln z − z . Thus we can extract the initial pressure. Now, the intercept is also equal
( )
τ
0
⎡ ⎤
⎣ ⎦ , and from this we obtain z∞ .
k
to ln ( z z0)( 1 e ) τ −
∞ − −
In the Kezdy-Swinbourne formalism, dividing equation [1] by equation [2] gives:
zt−z∞ z − z
t+
τ ∞
and ( 1 )
A plot of z t vs. zt+ τ gives a slope of
(for z∞ ).
Note that zt zt+ τ
with the 45 o line, this corresponds to z∞ .
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= e
kτ
z = z − e + z e
kτkτ t ∞ t+
τ
k
e τ k
(to extract k ) and an intercept of z ( 1 e ) τ
∞ −
= at t =∞ (equilibrium). So when the experimental data intersects
In general, for either method, you want to use a τ value that is 2 – 3 half lives

Half-lives:
One relevant parameter in the interpretation of kinetic data is called the half-life. The
[ A] = [ A]
/2.
half-life τ 1/2 is defined as the time when 0
Zero order reaction:
From the integrated rate law: [ A] = [ A] 0 − kt
The half-life is: [ A] 0 /2 = [ A] 0 − kτ1/2
So:
First order reaction:
[ A]
τ 1/2 =
2k
kt
From the integrated rate law: [ A] [ A] 0e
−
=
k
[ A] /2 [ A] e τ −
=
So: 1/2
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0
0 0
− ln 2 =− kτ
τ =
ln 2
k
Note that the first order half-life does not depend on [A]0.
Second order reaction – Type I:
From the integrated rate law:
1/2
1 1
=
[ A] [ A]
0
+ kt
2
[ A] 1
=
[ A]
+ kτ
Now: 1/2
0 0
So: 1/2
Pseudo-first order reaction:
1
τ =
[ A] k
ln 2 ln 2
Similar to the first order reaction: τ 1/2 = =
k' [ A] 0 k
Where k ' is the pseudo-first order rate constant and k is the second order rate
constant.
0
1/2

VI. Thermodynamic vs. kinetic control:
There will be a question on the exam regarding this material…
Consider the reaction:
When run under different conditions, different product ratios are obtained:
Why?
HBr
⎯⎯⎯→
Recall parallel reactions:
Conditions: 1-2 addition 1-4 addition
-80 o C 81% 19%
Room Temp 44% 56%
45 o C 15% 85%
k1
A⎯⎯→ B
k2
A⎯⎯→C In reality, however, these are really equilibrium processes:
A←⎯⎯ ⎯⎯→ B
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k1
k−1
k2
k−2
A←⎯⎯ ⎯⎯→ C
We can now express the rate laws for each component:
Br
−dA
[ ]
= k1[ A] + k2[ A] −k−1[ B] − k−2[ C]
dt
dB [ ]
= k1[ A] − k−1[ B]
dt
dC [ ]
= k2[ A] − k−2[ C]
dt
If we consider very short times (i.e.: [B] and [C] are so small that the reverse rates do
not occur), we need only consider the forward rates. We obtain the expressions for
[B] and [C] as a function of time:
+
Br

k [ A]
B = −e
1 0 [ ] 1
k1+ k2
− ( k1+ k2) t { }
This is from before. We also found that:
finally:
[ B] k
[ C] k
1 = , so
2
[ B]
= e
[ C]
k [ A]
k + k
2 0 − ( k1+ k2) t
and [ C] = { 1−e
}
[ B] Ae
=
[ C] Ae
−ΔEa/
RT
1 2
−Ea(
B)/ RT
−Ea(
C)/ RT
For short times, the ratio of products depends only on the difference in activation
energies! The product with the larger forward rate constant is called the kinetic
product, and its concentration is dominant for short times in the reaction.
Now consider the reaction at long (equilibrium) times. We do not have to solve the
rate laws, just look at the equilibrium conditions:
K
1
[ B]
[ C]
[ B] K1
= and K2
= so =
[ A]
[ A]
[ C] K
So the ratio depends on the relative equilibrium constants.
[ B] K e
[ C] K e
−ΔG(
B)/ RT
= 1 =
2
and
−ΔGC
( )/ RT
[ B]
= e
[ C]
D. Savin © 2009 PSC 480 Kinetics Supplement Page 30 of 33
2
−ΔΔG/
RT
At equilibrium, the ratio of products depends on the difference in Δ G values for each
product. The product with the lower Δ G is called the thermodynamic product, and
its concentration is dominant at long times in the reaction.

Since Ea(1− 2) < Ea(1−
4) , k(1− 2) > k(1−
4) , thus 1-2 addition product formed faster
initially. The 1-2 addition is the kinetic product.
But since ΔG(1− 2) >ΔG(1− 4) , the 1-4 addition is favored at equilibrium. The 1-4
addition is the thermodynamic product.
Remember – a faster reaction does not imply a lower Δ G !!
Also – one product can be favored both initially and at equilibrium (depending on the
relative activation energies and G
Δ ’s.
VII. Perturbation/Relaxation Kinetics
Imagine you want to study the kinetics of a reaction that reaches equilibrium in a very
short amount of time. The specific information of concentration vs. time may not be
experimentally accessible. In such an experiment, you can perform a perturbation
experiment, where the equilibrium conditions are altered by a sudden change in some
external variable such as temperature or pressure. Then, the rate of adjustment to the
new equilibrium concentrations is measured.
The key assumption is that the time required to apply the perturbation is much smaller
than the reaction time.
Consider the first order reaction:
A←⎯⎯ ⎯⎯→ B
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k1
k−1
−dA
[ ]
rate = = k1[ A] − k−1[ B]
dt
[ B]
eq
At equilibrium: = K1
[ A]
eq
Now we apply a perturbation (i.e.: T-jump) to the equilibrium system and monitor the
kinetics as the system relaxes to the new equilibrium.
We can express the concentrations of each component as the sum of the initial
equilibrium concentration and some change in concentration:
After the perturbation:
[ A] = [ A] +Δ
[ A]
eq

[ B] = [ B] +Δ [ B]
We now take these two concentrations and plug them into the rate expression above:
now:
( [ ] eq [ ] )
− d A +Δ A
rate = = k1 A +Δ A − k 1 B +Δ B
dt
( )
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eq
( [ ] eq [ ] ) − ( [ ] eq [ ] )
− d [ A] eq +Δ[ A] −dA [ ] eq dΔ[ A]
= − = k1[ A] eq + k1Δ[ A] −k−1[ B] eq −k−1Δ [ B]
dt dt dt
but we know that
dA [ ] eq
− = 0
dt
We can now represent the change in concentration for each component in terms of a
common factor Δ [ X ] =−Δ [ A] =Δ [ B]
. This comes from the stoichiometry, and the
choice to have the Δ [ A]
term negative is arbitrary, as long as the signs for Δ [ A]
and
Δ [ B]
are different. We now express our rate in terms of this new variable:
But we know that at equilibrium:
dΔ[ X]
rate = = k1[ A] eq −k1Δ[ X ] −k−1[ B] eq −k−1Δ [ X ]
dt
k [ A] = k [ B]
[forward rate equals reverse rate]
1 eq −1
eq
dΔ[ X]
rate = =− k1+ k−1 Δ [ X ]
dt
Now: ( )
We can solve this differential equation by separating variables:
dΔ[ X]
=− ( k1+ k−1) dt
Δ[
X ]
0
( k k )
−
Δ [ X] =Δ [ X] e
− 1+ 1
Define the relaxation time τ as the time required for
point in the decay). For this particular reaction:
1
τ =
k +
k
1 −1
[ ]
Δ[
X ]
e
0
Δ X = (i.e.: the 1/e

For other reactions:
Key points:
Mechanism 1/τ
A←⎯⎯→ ⎯ B
k1+ k−1 A B⎯⎯→ P
k [ A] eq + [ B] eq + k− + ← ⎯ 1( ) 1
+ + ← ⎯ ( [ ] eq[ ] eq [ ] eq[ ] eq [ ] eq[ ] eq )
+ ←⎯⎯→ ⎯ +
k1( [ A] eq + [ B] eq ) + k−1( [ P] eq + [ Q]
eq )
A B C ⎯⎯→ P
A B P Q
2A⎯⎯→ A
k A B + B C + A C + k− 1 1
4 k [ A] eq + k− ← ⎯ 2
1 1
(1) Start with rate laws
(2) Apply perturbation, express in terms of Δ [ X ] (from stoichiometry)
2
(3) All Δ [ X ] eq terms cancel out, all Δ [ X ] and all dΔ [ X]/ dt terms go to zero
dΔ[ X]
(4) You are left with = ( stuff ) Δ [ X ] (first order behavior)
dt
(5) Integrate and solve for 1/e point.
Note that we have not considered coupled equilibria..
D. Savin © 2009 PSC 480 Kinetics Supplement Page 33 of 33