A diamond is forever.
Is It So?
Consider:
Gibb’s Free Energy
?
C diamond  C graphite
ΔG = ∑ ΔGproducts - ∑ ΔGreactants
ΔG = ΔGgraphite - ΔGdiamond
ΔG = (0) - (3 kJ/mol)
ΔG = -3 kJ/mol
Look quick, before it
turns into graphite.
While it’s true her diamond is spontaneously
turning into graphite before her eyes, it’s
happening very slowly. Don’t hold your
breath waiting to see any change. It takes
billions of years.
While thermodynamics answers the
question as to whether or not a reaction or
event is spontaneous, it DOES NOT tell how
fast a reaction goes.
This is what kinetics does....describes
the rate of the reaction.
Chemical Kinetics are about the only tool a chemist has
to probe the actual mechanism of a chemical reaction.
It is the study of reaction rates.
Reaction Rates
Reaction Rate: The slope
of a Concentration vs
Time graph.
NO2  NO + O2
Reaction Rates
Factors affecting reaction rates:
•The nature of the reaction mechanism
•The concentration of the reactants
•The temperature at which the reaction occurs
•The presence of a catalyst
•The surface area of solid or liquid reactants or
catalysts
Red  Blue
Rate Laws describe reaction
rates mathematically
d A

 k A
dt
Rate = k[reactant 1]m [reactant 2]n…
Rate Laws
To simplify the rate laws, we assume
conditions where only the forward reaction is
important. This produces rate laws that only
contain reactant concentrations.
There are two forms of the rate law…
Differential rate laws: Shows how
the rate depends on concentration.
Sometimes called just the “rate law”.
Integrated rates laws: Shows how the
concentration depends on time.
The choice of which rate law to use depends on the
type of data that can be collected conveniently and
accurately. Once you know one type, the other can be
calculated.
Integrated
Differential
The choice of which rate law to use depends on the
type of data that can be collected conveniently and
accurately. Once you know one type, the other can be
calculated.
Integrated
Differential
This of course requires
the use of my calculus.
Other Stuff to Know:
Reaction Order
The differential rate law for most reactions has
the general form….
Rate = k[reactant 1]m [reactant 2]n…
The exponents m and n are called reaction orders.
Their sum (m + n) is called the overall reaction order.
Other Stuff to Know:
Reaction Order
Usually the values of the reaction order must
be determined experimentally, they cannot be
predicted by looking at the chemical reaction.
In most rate laws, reaction orders are 0, 1, or
2. However, occasionally they are a fraction or even
negative.
The rate of a reaction depends on concentration;
however the rate constant does not.
Finding The Differential Rate
Law
The most common method for experimentally
determining the differential rate law is the
method of initial rates.
In this method several experiments are run
at different initial concentrations and the
instantaneous rates are determined for each at
the same value of time (as near t = 0 as possible)
Using Initial Rates to Determine
the Form of the Rate Law
Example:
A + B  C
Exp #
[A]
[B]
Initial Rate (M/s)
1
.100M
.100M
4x10-5
2
.100M
.200M
4x10-5
3
.200M
.100M
16x10-5
From this data, find the form of the rate law..
Rate = k[A]m[B]n
Exp #
[A]
[B]
Initial Rate (M/s)
1
.100M
.100M
4x10-5
2
.100M
.200M
4x10-5
3
.200M
.100M
16x10-5
Rate = k[A]m[B]n
Rate 2 4 105 k[.100]m [.200]n



5
m
Rate1 4 10
k[.100] [.100]n
[.200]n
n
1
n
1

2
n=0
[.100]
[B]0 = 1
Exp #
[A]
[B]
Initial Rate (M/s)
1
.100M
.100M
4x10-5
2
.100M
.200M
4x10-5
3
.200M
.100M
16x10-5
5
Rate3 1610
k[.200] [.100]



5
m
n
Rate1 4 10
k[.100] [.100]
m
[.200]
4
[.100]m
42
m
m
Rate = k[A]2[B]0 = k [A]2
n
m=2
Exp #
[A]
[B]
Initial Rate (M/s)
1
.100M
.100M
4x10-5
2
.100M
.200M
4x10-5
3
.200M
.100M
16x10-5
Now, solve for k…
rate
4 105
3
k


4

10
2
0
2
[ A] [ B]
[.100]
Rate = 4x10-3 [A]2
First Order Reactions
d A
 k A
Rate = 
dt
Integrating, etc. leads to…
[A] = [A]o e-kt
Ln [A]t – ln [A]0 = -kt
 At  
   kt
ln 
  A0  


Derive the Integrated First Order Rate Law:
Given a first order differential rate law…
[ A]
Rate 
 k [ A]
t
That is, the rate of change of the concentration of reagent A is
proportional to how much A there is.
Let x = [A]
(the concentration of A), Then…
x
  kx
t
The minus sign is because the value of [A], or x, is decreasing.
Transforming to calculus notation…
dx
  kx
dt
Separate the variables…
dx
  k dt
x
Integrate both sides…
1
 x dx    k dt
1
 x dx    k dt
Integrating…
Solving for x…
ln x  kt  C
ln x
e
e
 kt  C
 kt C
xe e
C  kt
xe e
C  kt
xe e
Now use initial conditions to determine C.
At t = 0, x = [A]o
[A]o = initial concentration of A
[ A] o  eC e  k ( 0 )  eC
ln [ A] o  ln e
C
C  ln [ A]o
Recall that..
C  kt
xe e
x [ A]  [ A]o e
 kt
eC  [ A]o
x [ A]  [ A]o e
Take ln’s of both sides…
ln [ A]  ln[ A] o e
 kt
 kt

 ln [ A]o  ln ekt
So…
ln [ A]  ln [ A] o   kt 
ln [ A]  ln[A]o  kt
First Order Reactions
d A
Rate  
 k A
dt
Ln [A]t – ln [A]0 = -kt
 At  
   kt
ln 
  A0  


These equations can be used to determine …
•The concentration of a reactant remaining at any
time after the reaction has started.
•The time required for a given fraction of a sample to
react.
•The time required for a reactant concentration to
reach a certain level.
Half Life (First Order Process)
Half Life of a Reaction
The half life of a reaction, t½ , is the time required
for the concentration of a reactant to drop to one half
of its initial value.
  At  
ln 
 A  
   kt
 0 
 12 A0  
   kt 1
ln 
2
 A0  
ln (½) = -kt½
ln 12 
t1  
2
k
T½ = 0.693/k
Half life for a first order rate law is independent of
the initial concentration of reactant. Thus, the half life is
the same at any time during the reaction.
Half Life for a First Order Reaction
(Garden Variety Half Life – Radioactivity)
First order reaction
example
The first order rate constant for the decomposition of a
certain insecticide in water at 12oC is 1.45 g/mLyr. A
quantity of this insecticide is washed into a lake on June
1, leading to a concentration of 5.0x10-7 g/ml water.
Assume that the effective temperature of the lake is
12oC.
a. What is the concentration of the
insecticide on June 1 of the following year?
b. How long will it take for the concentration
to drop to 3.0x10-7 g/ml?
a. What is the concentration of the insecticide on
June 1 of the following year?
Ln [A]t – ln [A]0 = -kt
Ln [A]t = -kt + ln [A0]t=0
ln [insecticide] t=1 year = - (1.45g/mLyr)(1 yr) + ln (5.0x10-7
g/ml)
ln [insecticide] t=1 year = - (1.45g/mL) + (-14.51g/mL)
ln [insecticide] t=1 year = -15.96 g/mL
[insecticide]
t=1 year
[insecticide]
t=1 year
= e-15.96
= 1.2x10-7 g/mL
b. How long will it take for the concentration of the
insecticide to drop to 3.0x10-7 g/ml?
ln [A]t = -kt + ln [A0]t=0
ln (3.0x10-7 g/ml) = - (1.45 yr-1)(t) + ln (5.0x10-7 g/ml)
t = -[ln (3.0x10-7g/ml) – ln (5.0x10-7g/ml)]/1.45yr-1
t = - [-15.02 + 14.51]/1.45yr-1
t = 0.35 yr
Second Order Reactions
A second order reaction is one whose rate
depends on the reactant concentration raised to the
second power, or on the concentrations of two different
reactants, each raised to the first power. For a
reaction that is second order in just one reactant, A,
the rate law is given by…
Rate = k[A]2
Second Order Reactions
Rate = k[A]2
A
2
 k A
t
Integrating and stuff leads to the Integrated
Rate Law…
1
1
 kt 
At
A0
Second Order
Half Life’s
1
1
 kt 
At
A0
Setting [A] = ½[A]o and solving for time gives…
1
t1 
2
k A0
Unlike the half life of first order reactions, the
half life of a second order reaction is dependent on
the initial concentration of the reactant.
Half Life for a Second Order Reaction
Zero Order Reaction
A zero order reaction is one whose rate depends on
the reactant concentration raised to the zero power, or in
other words, it does not depend on the concentration of
the reactant as long as some reactant is present. For the
reaction that is zero order in just one reactant, A, the
rate law is given by…
Rate = k[A]0
Integrating…
Solving for t½
[A]t = -kt + [A]0
T½ = [A]0/2k
Half Life for a Zero Order Reaction:
(Rate doesn’t depend on concentration)
Rate Laws: A Summary
To simplify the rate laws, we assume conditions
where only the forward reaction is important. This
produces rate laws that only contain reactant
concentrations.
There are two types of rate laws…
Differential rate laws: Shows how the rate
depends on concentration.
Integrated rate laws: Shows how
concentration depends on time.
Rate Laws: A Summary
The most common method for experimentally
determining the differential rate law is the method of
initial rates. In this method several experiments are run
at different initial concentrations and the instantaneous
rates are determined for each at the same value of time
(as near t=0 as possible).
The point is to evaluate the rate before the
concentrations change significantly from the initial values.
From a comparison of the initial rates and the initial
concentrations the dependence of the rate on the
concentrations of various reactants can be obtained – that
is the order of each reactant can be determined.
Rate Laws: Summary
To experimentally determine the integrated rate
law for a given reaction, concentrations are measured at
various values of time as the reaction proceeds. Then
the job is to see which integrated rate law correctly fits
the data. Typically this is done visually, by ascertaining
which type of plot gives a straight line.
Once the correct straight line plot is found, the
correct integrated rate law can be chosen and the value
of the rate constant, k, can be obtained from the slope
of the plot. Also, the differential rate law for the
reaction can be calculated.
All the Good Stuff
Reaction
Order
Zeroth
Differential
Rate Law
d A

k
dt
First

d A
 k A
dt
Integrated Rate
Law
A  Ao  kt
A  Ao ekt
Plot for
Straight Line
A vs t
lnA vs t
Second
Slope of
Plot
-k
-k
k

d A
 k A 2
dt
A 
A
1  ktAo
1
vs t
A
We can use this information to experimentally find
the rate law for a reaction.
Reaction
Order
Zeorth
Differential
Rate Law
d A

k
dt
First

d A
 k A
dt
Integrated Rate
Law
A  Ao  kt
A  Ao ekt
Plot for
Straight Line
A vs t
lnA vs t
Second
Slope of
Plot
-k
-k
k

d A
 k A 2
dt
A 
A
1  ktAo
1
vs t
A
The gas phase decomposition of NO2 is studied at
383oC giving the following data.
NO2  NO + O2
Time (sec)
50
100
150
200
250
Conc. NO2
Molar
0.1000
0.0170
0.0090
0.0062
0.0047
Original Data
Time
Molar
Conc.
Graph the ln (concentration) vs Time. It
the plot is a straight line, the reaction is
first order.
ln Concentration vs Time
ln Conc.
0.0000
0
100
150
200
250
0.0170
0.0090
-2.3026
-4.0745
-4.7105
10
15
-1.0000
ln concentration
50
0.1000
5
-2.0000
-3.0000
-4.0000
-5.0000
0.0062
-5.0832
-6.0000
Time (sec)
0.0047
Graph
-5.3602
20
25
Original Data
Time
concentration-1 vs Time
Conc.-1
250.0000
0.1000
10.0000
0.0170
58.8235
150
200
250
0.0090
111.1111
0.0062
161.2903
Conc
100
y = 10.16x + 9.1984
200.0000
-1
50
Molar
Conc.
Graph the 1/(concentration) vs Time. It
Reaction
constant
k reaction
= 10.16 is
the plot
is a straight
line, the
second order.
150.0000
100.0000
50.0000
0.0000
0
5
10
15
20
Time
0.0047
212.7660
Straight line  Second
Order Reaction
25
Remember that the rate equation between two
substances A and B looks like this…
Rate = k[A]m[B]n
The equation shows the effect of changing the
concentrations of the reactants on the rate of the
reaction. What about all the other things like
temperature and catalysts, for example which also change
rates of reaction? Where do these fit into this equation?
These are all included in the so-called rate
constant – which is only actually constant if all you are
changing is the concentration or the reactants. If you
change the temperature or the catalyst the rate
constant changes. These changes of the rate constant
are shown mathematically in the Arrhenius Equation.
Svante August Arrhenius
Svante August Arrhenius was a Swedish
physical chemist best known for his theory that
electrolytes, certain substances that dissolve in
water to yield a solution that conducts
electricity, are separated, or dissociated, into
electrically charged particles, or ions, even when
there is no current flowing through the solution.
In 1903 he was awarded the Nobel Prize
for Chemistry.
1859-1927
The Arrhenius Equation
k  Ae

Ea
RT
K = rate constant
A = frequency factor
Ea = activation energy
R = the gas constant (8.314 J/mol·K)
T = temperature in Kelvin
The Arrhenius Equation
Eactivation

RT
k  Ae
Or the logarithmic form…
Eactivation
ln k  ln A 
RT
The expression…
E activation

e RT
For reasons that are beyond the scope of any course at
this level, this expression counts the fraction of the
molecules present in a gas which have energies equal to or in
excess of the activation energy at a particular temperature.
Eactivation

RT
k  Ae
The frequency factor, A is sometimes called
the pre-exponential factor of the steric factor.
It is a term which includes factors like the
frequency of collisions and their orientation. It
varies slightly with temperature, although not much.
It is often taken as constant across small
temperature ranges.
Using the Arrhenius Equation:
The effect of a change of temperature.

k  Ae
Eactivation
RT
You can use the Arrhenius equation to show the
effect of a change of temperature on the rate constant –
and therefore on the rate of the reaction.
If the rate constant doubles, for example, so will
the rate of the reaction.
Example: What happens to the rate of a reaction if the
temperature increases from 20oC to 30oC?
The frequency factor, a, is approximately constant for
such a small temperature change. We need to look at how e-(Ea/RT)
changes --- the fraction of molecules with energies equal to or in
excess of the activation energy.
Assume an activation energy of 50 kJ/mol. In the
equation, we have to write that as 50,000 J/mol. The value of the
gas constant R is 8.31 J/K·mol
Now raise the temperature just a
little bit to 30oC (303k)
E
 activation
e RT
e
 1.2  109

50,000
8.31293
e

E activation
RT
e
 2.4  109

50,000
8.31303
At 20oC
e

E activation
RT
e
 1.2  109

At 30oC
50,000
8.31293
E
 activation
e RT
e

50,000
8.31303
 2.4  109
The fraction of the molecules able to react has
almost doubled by increasing the temperature by 10oC.
That causes the rate of the reaction to almost double.
This is the basis for the old rule of thumb
that a reaction rate doubles for every 10oC rise in
temperature.
At the higher temperature, more molecules have energy
greater than the activation energy Ea.
Catalysts:
A substance that changes
the rate of a reaction without being
consumed in the reaction.
•Provides an easier way to react.
Cattleist?
•Lowers activation energy
•Enzymes are biological catalysts.
•Inhibitor: A substance that decreases the rate
of reaction (a negative catalyst)
The effect of a catalyst is to lower the activation
energy.
Lower activation energy means more molecules will react.
Uncatalyzed
Energy
Catalyzed
Energy
The effect of a catalyst
A catalyst provides an alternate reaction mechanism, or
route for the reaction. This alternate route necessarily has a lower
activation energy. The overall effect of a catalyst is to lower the
activation energy.
Continuing our example of a reaction with an
activation energy of 50,000 J/mol… What is the effect
on the rate of lowering the activation to 25,000 J/mol?
With Catalyst
E activation

e RT
25, 000

 e 8.31293
 3.5  1 05
Without Catalyst
E activation

e RT
e
9
 1.2  10
No wonder catalysts speed up reactions!

50,000
8.31293
The Collision Model
Molecules must collide to react. The greater
the number of collisions occurring per second, the
greater the reaction rate.
Only a small fraction of collisions actually
lead to a reaction.
In 1888 the Swedish chemist Svante Arrhenius
suggested that molecules must possess a certain minimum
amount of energy in order to react. In order to react,
colliding molecules must have a total kinetic energy equal
to or greater than some minimum value. This minimum
energy is called the activation energy (Ea). The value of Ea
varies from reaction to reaction.
Factors Influencing
Reaction Rates
The rate of a given reaction can be affected by:
•
•The concentration of the reactants (for gases,
•
•
The physical states of the reactants
pressure is equivalent to concentration)
The reaction temperature
The presence of a catalyst
The greater the concentration, the
greater the frequency of collisions.
The Greater the Temperature the
Greater the Frequency of Collisions.
An increase in temperature generally increases the
rate of a reaction.
According to kinetic theory, at higher temperature
the atoms and molecules move faster.
1. They collide more frequently
2. They collide with greater energy and each
collision has a greater chance of being successful.
Note: The activation energy does NOT
get any lower!
Not all collisions produce a reaction
Energy Diagrams
Reactants
Intermediaries – Activated
Complexes
Products
Molecularity
Chemical reactions take place as a result of collisions
between molecules. Each type of collision in a given reaction
is called a single event or an elementary step.
The number of molecules that participate as reactants
in an elementary step defines the molecularity of the step.
Unimolecular: A single molecule is involved in the
elementary step
Bimolecular: Two molecules are involved
Termolecular: Three molecules are involved. Termolecular
steps are far less probable than unimolecular or bimolecular
processes and are rarely encountered.
The chances that four or more molecules
will collide simultaneously with any
regularity is extremely remote,
consequently, such collisions are never
proposed as a part of a reaction
mechanism.
Example:
The reaction between NO2 and CO has the
overall reaction:
NO2 + CO  NO + CO2
A study of the kinetics of this reaction revealed the
rate law for the reaction is.
Rate = k[NO2]2
This Rate Law requires that the slow step of
the reaction involves a collision between two NO2
molecules. How can this be a step in the seemingly
simple reaction above?
NO2 + CO  NO + CO2
Further study of this reaction established that
two NO2 molecules can react as follows…
NO2 + NO2  NO3 + NO
NO3 is a highly reactive material which is capable
of transferring an oxygen atom.
NO3 + CO  NO2 + CO2
2NO2 + NO3 + CO  NO2 + NO3 + NO + CO2
NO2 + CO  NO + CO2
NO2 + CO  NO + CO2
The first equations sets the rate law, so it must be
the slow one.
NO2 + NO2  NO3 + NO
slow
NO3 + CO  NO2 + CO2
fast
NO2 + CO  NO + CO2
Remember, the rate law only provides information
about the slowest reaction in the mechanism.
Experimental evidence for reaction mechanisms:
Consider the reaction of methyl acetate with water….
Methyl Acetate
+
water 
Acetic Acid + Methanol
HO
H
H
H
H-C-C-O-C-H + H2O  H-C-C-OH + H-C-OH
H
H
H
H
The methyl acetate molecule can break in one of two places
to complete the reaction…
Methyl Acetate
+
water 
Acetic Acid + Methanol
HO
H
H
H
H-C-C-O-C-H + H2O  H-C-C-OH + H-C-OH
H
H
H
H
HO
H-C-CH
H
O-C-H
H
H
O-H
-OR-
Methyl Acetate
+
water 
Acetic Acid + Methanol
HO
H
H
H
H-C-C-O-C-H + H2O  H-C-C-OH + H-C-OH
H
H
H
H
HO
H-C-CH
H
O-C-H
H
H
O-H
-OR-
HO
H
H-C-C-O-
-C-H
H
H
H
O-H
How can we determine experimentally which is correct?
Use a radioactive isotope of oxygen,
O18 To make up the water
H-O18
H
HO
H-C-C-O18-H
H
Methyl Acetate
+
water 
Acetic Acid + Methanol
HO
H
H
H
H-C-C-O-C-H + H2O  H-C-C-OH + H-C-OH
H
H
H
H
HO
H-C-CH
H
O-C-H
H
H
O-H
-OR-
HO
H
H-C-C-O-
-C-H
H
H
H
O-H
Elementary Steps
Experimental studies of reaction mechanisms
begin with rate measurements. Next this data is
analyzed to determine the rate constant and order of
the reaction. The rate law is then written, and finally
a plausible mechanism for the reaction in terms of
elementary steps is formulated.
The elementary steps must satisfy two requirements…
1.
The sum of the elementary steps must give the overall
balanced equation for the reaction.
2. The rate-determining step, which is the slowest step in the
sequence of steps leading to product formation, should
predict the same rate law as is determined experimentally.
Rate Laws of Multistep
Mechanisms
•The slowest elementary step is the rate determining
step.
•The rate determining step governs the rate law for the
overall reaction.
•For elementary steps, the exponent in its rate are the
same as the coefficients of the reactants in the chemical
equation. Thus, the rate law for an elementary process can
be predicted from the chemical equation for the process.
Rate Laws of Elementary Steps
A  Products
Rate = k[A]
A+A  Products
Rate = k[A]2
A+B  Products
Rate = k[A][B]
A+A+A  Products
Rate = k[A]3
A+A+B  Products
Rate = k[A]2[B]
A+B+C  Products
Rate = k[A][B][C]
Measuring the
rate of a
reaction.
Formulating the
Rate Law
Postulating a
reasonable
reaction
mechanism
Sequence of steps in the study of a
reaction mechanism
Example:
The decomposition of Hydrogen
Peroxide is facilitated by iodide
ions.
The overall reaction is…
2H2O2  2H2O + O2
By experiment, the rate law is
found to be…
rate = k[H2O2][I-]
Thus the reaction is first order
with respect to both H2O2 and I-.
Decomposition of
Hydrogen Peroxide
2H2O2  2H2O + O2
rate = k[H2O2][I-]
You can see that H2O2 decomposition does not occur in a
single elementary step corresponding to the overall balanced
equation.
If it did, the reaction would be second order in H2O2 (as a
result of the collision of two H2O2 molecules). What’s more, the
I- ion, which is not even part of the overall equation, appears in
the rate law expression.
We can account for the observed rate law by assuming
that the reaction takes place in two separate elementary steps,
each of which is bimolecular:
step 1:
H2O2 + I-  H2O + IO-
step 2:
H2O2 + IO-  H2O + O2 + I-
Slow
step 1:
H2O2 + I-  H2O + IO-
step 2:
H2O2 + IO-  H2O + O2 + I-
The first step must be the rate determining step, because
it matches the rate law. Thus, the rate of the reaction can be
determined from the first step alone:
rate = k[H2O2][I-]
Note that the IO- ion is an intermediate because it does not
appear in the overall balanced equation. Although the I- ion also does
not appear in the overall equation, I- differs from IO- in that the
former is present at the start of the reaction and at its completion.
The function of I- is to speed up the reaction --- that is, it is a catalyst.
Another
Example:
Consider the Reaction…
2NO2Cl  2NO2 + Cl2
This is found to be a first order reaction, it’s experimentally
determined rate law is…
rate = k[NO2Cl]
Could the overall reaction occur in a single step by the
collision of two NO2Cl molecules? This would lead to…
rate = k[NO2Cl]2
Chemists believe the actual mechanism is…
NO2Cl  NO2 + Cl
Slow
NO2Cl + Cl  NO2 + Cl2
Fast
Still Another Example:
Consider the reaction…
2NO + 2H2  N2 + 2H2O
Experimentally, the rate law has been found to be…
rate = k[NO]2[H2]
A chemically reasonable mechanism that yields the correct
form for the rate law is…
2NO + H2  N2O + H2O
(slow)
N2O + H2  N2 + H20
(fast)
However, there is a serious flaw in the proposed mechanism --it has a elementary process that involves the simultaneous collision
between three molecules, two of which must be NO and one H2.
Consider the reaction…
2NO + 2H2  N2 + 2H2O
An alternative proposal is…
2NO ↔ N2O2
(fast)
N2O2 + H2  N2O + H2O
(slow)
N2O + H2  N2 + H2O
(fast)
Since the second step is rate determining, the rate for the
reaction should be…
rate = k[N2O2][H2]
However, the experimental rate law does not contain the
species N2O2, an intermediate.
2NO ↔ N2O2
(fast)
N2O2 + H2  N2O + H2O
(slow)
N2O + H2  N2 + H2O
(fast)
The first elementary step is an equilibrium reaction, so
consider both forward and reverse reactions.
Rate (forward reaction) = kf[NO]2
Rate (reverse reaction) = kr[N2O2]
If, however, we view the first equation as a dynamic
equilibrium reaction, then the rate of the forward and reverse
reactions are equal. Then…
Kf[NO]2 = kr[N2O2]
[N2O2] = kf/kr[NO]2
Kf[NO]2 = kr[N2O2]
[N2O2] = kf/kr[NO]2
Substituting into the proposed rate law (rate = k[N2O2][H2])
gives….
rate = k (kf/kr)[NO]2[H2]
Or
rate = k’ [NO]2[H2]
The rate law derived from the proposed mechanism
matches the experimental rate law, thus the three step
mechanism does appear to be reasonable on the basis of
kinetics.