Chapter 18 Chemical Kinetics: Mechanisms
In the last chapter we went through the mechanics of how you extract rate constants
and order parameters from experimental data. In this chapter we will ge tthe theory of
what these numbers mean
18-1 Mechanisms
Up to now we have worked with balanced chemical equations. These equation
show the stoichiometric relationships between products and reactants, but it says
nothing about how fast the reaction goes, or how the reaction actually occurs.
This is what we will try to do now.
The first assumption we will put into our theory of kinetics is that two molecules
must collide for a reaction to occur between them.
For now let’s define an elementary step as a chemical reaction that describes
the simple collision between two molecules. I will come back with a better
definition later
We find that many chemical reactions appear to be more complicated than a
simple single collision, and seem to involve one or more elementary steps, thus
we will describe a chemical reaction as a series of elementary steps called a
reaction mechanism
Key concept:
A reaction mechanism is a series of elementary steps that add up to the
balanced stoichiometric equation
The reaction mechanism ends up becoming a step-by-step description of what
happens in a chemical reaction
Let’s start with the reaction NO2(g) + CO(g) 6 NO(g) + CO2(g)
The simplest assumption is that NO2 collides with CO, and one O is transferred
from the NO2 to the CO in this reaction to make it complete.
Reaction mechanism
Step 1: NO2(g) + CO(g) 6 NO(g) + CO2(g)
If this were the case, then the reaction rate would depend on the concentrations
of both NO2 and CO because both molecules are involved in the reaction. Thus
rate = k[NO2][CO].
Kinetic analysis of the experimental data tells us this is wrong. The experimental
data says that rate = k[NO2]2 and so is not consistent with the real data
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Let’s try another mechanism
Reaction Mechanism
Step 1: NO2(g) + NO2(g) 6NO3(g) + NO(g)
Step 2: NO3(g) + CO(g) 6 NO2(g) + CO2(g)
Key Concept:
Test 1 of a reaction mechanism: Does the sum of the elementary steps = the
stoichiometric equation for the reaction?
Test1 for this reaction
Sum of step 1 and step 2
= NO3(g) + CO(g) + NO2(g) + NO2(g) 6NO3(g) + NO(g)+NO2(g) + CO2(g)
=CO(g) + NO2(g) 6 NO(g) + + CO2(g) Passes check 1!
In this mechanism NO3 was created in one step and destroyed in another, so it
never appears in the final balanced reaction.
Key Concept:
An intermediate species, or intermediate, is a molecule that is produced in one
elementary step and destroyed in subsequent step and therefore does not appear in
the overall reaction.
Intermediates are usually unstable, reactive compounds. While unstable,
sometimes they can be detected in a reaction mixture, even it they exist only for
a microsecond.
Back in the last chapter I said that you cannot write a rate law equation from the
stoichiometric equation. Why ? Because stoichiometry only relates mole ratios
of products and reactants to each other and has nothing to do with kinetics or the
chemical mechanism.
However since an elementary step describes a molecular collision and not a
stoichiometeric relationship, you CAN write the rate law from the molecularity of
an elementary step.
An elementary step that involves the collision between two molecules is called a
bimolecular step. An elementary step that involves a change in a single
molecule is called a unimolecular step
Let’s go back to our definition of elementary step and try to improve it a bit
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Key Concept:
An elementary step is a single step in a reaction mechanism that describes how
one molecule is transformed or how two molecules collide to produce one or more new
molecules. The reaction rate law of an elementary step may be written from the
molecularity of that step
Example problem Possible clicker question?
The reaction 2NO2 + F2 6 2NO2F
Is thought to undergo a 2 step mechanism:
Step 1: NO2 + F2 6 NO2F + F
Step 2: F + NO2 6 NO2F
Does the sum of the elementary steps = stoichiometric equation? (YES)
Does this reaction have an intermediate? (YES F)
Write the reaction rate laws for each step
Step 1 rate = k[NO2][F]
Step 2 rate = k[F][NO2]
18-2 Activation Energy 18-3 Arrhenius Equation
We haven’t finished with our reaction mechanism yet, but I want to cover the
next subject first. In this discussion I will mix parts of 18-2 and 18-3 together
So now for our model of kinetics we say a reaction involved a collision between
molecules or a transformation within a molecule, and for each elementary step
we can write a reaction rate equation:
rate = k [ ].[ ]...
Where each concentration represents a molecule that takes part in that step
But what does the k mean? The k has to do with how fast the reaction occurs.
With that in mind, remember in the section on gases we could calculate how
quickly one molecule collides with another molecule?
In a 1L mixture of 2 gases at 1 bar and 25oC a molecule has about 1031
collisions/sec
If every collision in a gas resulted in a reaction then
rate of reaction = 1031 collision x 1 mol = 107 mol/L@sec
L@sec
6.022x1023
In fact almost no reactions occur at this high a rate, and most reaction occurs at
rates 100' s or 1000's of time slower. Why?
4
Think demolition derby. Do all the collisions in a demolition derby end with one
car dead and the other one driving away in fine shape? Does a driver care about
where he hits the other car with his car?
No and Yes. A slight bump between two cars won’t put either one out of
commission, and the drivers of demo derby cars are always trying to run their
rear end into the other car.
Bottom line, the energy of the collision and the orientation of the colliding objects
makes a big difference in whether the collision is successful or not. The same is
true in our collisional theory of chemical kinetics. Not only do we have collisions,
but the collision has to have enough energy and be in the right orientation before
a reaction will occur. Figure 18.1 if you want it
rate of reaction = Collisional
Frequency
X Fraction of collisions X fraction of collisions
With right orientation
With enough energy
Looking at individual terms
Collisional frequency % [A][B]....The higher the concentrations, the more
collisions. This term will also depend on T, the higher the T, the faster the
molecules move, and the more collisions you will have
Energy of Collisions - The minimum energy required to cause a reaction
between two molecules is called the activation energy (Ea). Figure 18.2
shows the energy distribution of molecules at two different temperatures.
As you can see as the T increases, the number of molecules that exceed
a fixed activation energy increases
Fraction of collisions with right orientation - a factor that says waht fraction of the
collisions are oriented correctly
Putting these factors into a single equation
k = zpe-Ea/RT
where z is the collisional frequency
p is the orientation factor, called the steric factor
Ea is the activation energy
R is the gas constant in Joules
T is the temperature in K
e-Ea/RT puts in the temperature dependence of the # of molecules with
E > Ea
Since z and p do not vary significantly with T they are often lumped together into
a single factor A, in a equation called the Arrehenius equation
k =Ae-Ea/RT
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k then, varies with temperature through an exponential function. Let’s see if we
can come up with an equation that shows how k varies with T that doesn’t have
that nasty e-Ea/RT term
Taking the ln of both sides
ln(k) = ln(A) + ln(e-Ea/RT)
ln(k) = ln(A) -Ea/RT
At this point, if you have k at several temperatures you can determine Ea by
plotting ln(k) vs 1/T. Why? Remember your line of best fit?
Y = mX + b
Y = ln(k)
X =1/T
m = slope = -Ea
b = intercept = ln(A)
On a test you won’t have time to plot a bunch of points, so I want to show you a
simpler two point equation
Now let’s get k1 at T1 and k2 at T2
ln(k1) = ln(A) -Ea/RT1 ;
ln(k2) = ln(A) -Ea/RT2
Now subtract the first equation from the second
ln(k2) - ln(k1) = ln(A) -Ea/RT2 - ln(A) -Ea/RT1
dropping out the ln(A) terms since they are equal
ln(k2) - ln(k1) = -Ea/RT2 - (-Ea/RT1)
Grouping the -Ea/R terms together
ln(k2) - ln(k1) = -Ea/R(1/T2 - 1/T1)
Using the properties of ln
ln (k2/k1) = -Ea/R (1/T2 - 1/T1)
Now getting rid of the - sign on the right side by switching T1 and T2 around
ln (k2/k1) = Ea/R (1/T1 - 1/T2)
Key Equations
k = zpe-Ea/RT
k =Ae-Ea/RT
You probably won’t do any calculations with these two equations, but you need
to know what they mean
ln(k) = ln(A) -Ea/RT
ln (k2/k1) = Ea/R (1/T1 - 1/T2) -or- ln (k2/k1) = Ea/R[ (T1-T2)/(T1T2)
These equations you will calculate with
6
Sample calculations:
Find the activation energy Ea for the reaction
NO + O36 NO2 + O2
Given the data in the following table:
T(K)
195
230
260
298
369
k(L/mol s)
1.08x109
2.95
5.42
12.0
35.5x109
What we really want is a plot of ln(k) vs 1/T so:
1/T
ln(k)
5.13x10-3
20.65
4.35
21.81
3.85
22.41
3.36
23.21
-3
2.71x10
24.29
Plotting in excel
slope = -1489 = -Ea/R
-Ea/R = -1489
Ea = 1.489x103 (R) = 1.489x103 (8.3145) = 12.4 kJ/mol
For shorter problems we can derive a simple equation for 2 point determination of
activation energy
7
Using the 2 point equation
ln(k2/k1) = Ea/R (1/T1 -1/T2)
And making the first point T2 and the last point T1 (it doesn’t matter which is
which)
ln(1.08x109/35.5x109) = Ea/8.3145(1/369 -1/195)
ln(1.08/35.5) = Ea/8.3145(.00271 -.00513)
-3.493=Ea/8.3145 ( -.00242
Ea = 3.493(8.3145)/.00242 = 12 kJ
about the same, just some roundoff error on the calculation
The meaning of activation energy
Now that you know how to find Ea from experimental data, let’s try to get a better
understanding for what it means
Earlier I said that the activation energy is the minimum energy required for a
properly oriented collision to create a chemical reaction. It is useful to think of an
activation energy in terms of an activation energy diagram Figure 18.3 or 18.4 on
the board
In this diagram we have the energy of the reactants on the left side, the energy
of the products on the right side, and you can think of the activation energy as
the height of a hill between the two that has to be overcome before a reaction will
occur
Example problems clicker type
Determine if figure 18.3 is an exothermic or endothermic reaction
Determine if figure 18.4 is exothermic or endo thermic
If the reaction in 18.3 has a ÄH of -10 and an activation energy of +30, what is
the activation energy for the REVERSE reaction
If the reaction in 18.4 has a Ea of 40 in the forward direction and an Ea of 20 in
the reverse direction, what is the ÄH for the reaction in the forward direction
The top of the energy hump of an activation energy diagram has a useful
physical interpretation. It is often interpreted as the energy of a transition state
or activated complex
For example the reaction
CH3Cl(aq) + OH- (aq) 6 CH3OH(aq) + Cl- (aq)
has kinetics where
rate = k [CH3Cl][OH-]
that says there is a key elementary step where the CH3Cl molecule collides with
an OH- ion
8
One can theorize that this goes through an intermediate as shown in the
mechanism shown on page 656
The square bracket with the funky superscript indicates a proposed intermediate
or activated complex that occurs at the peak of the activation energy hill. Since
this is at a higher energy than either the product or reactant, it is not stable, so
you cannot isolate it, but sometimes you can gather data (usually
spectroscopically) that implies it is there
Key Definitions
Be able to define these terms that you have seen in these two sections:
Activated complex or transition state
activation energy
steric factor
frequency factor
activation energy diagram
18-4 Rate-Determining Step
Now that we can interpret the kinetics of an elementary step, Let’s return to
putting the elementary steps together to come up with a reaction mechanism
Returning to the reaction
NO2(g) + CO(g) 6 NO(g) + CO2(g)
We had learned that it wasn’t a simple 1 step reaction and had proposed the
following mechanism:
Step 1: NO2(g) + NO2(g) 6NO3(g) + NO(g)
Step 2: NO3(g) + CO(g) 6 NO2(g) + CO2(g)
The sum of the elementary steps = the stoichiometric equation so it passed
check 1!
Now on to check 2, does it pass experimental kinetic data?
Well we have 2 elementary steps
The first would have rate = k[NO2]2
The second would have rate = k[NO3][CO]
Which rate should we focus on?
Key Concept:
In general, in any reaction mechanism, the rate determining step is the slowest
step in a reaction, and that single slow step determines the overall kinetic of the
reaction.
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If you have a hard time with this concept, think of driving to Rapid through a one
lane construction zone; if you are stuck behind granny doing 40, you won’t get to
Rapid any faster than her, even if you do have a hot car!
So if we propose that step 1 is the slow, rate determining step of our reaction
mechanism. Our mechanism agrees with experimental data, and our proposed
mechanism is consistent with experimental data
Example problem Possible clicker question?
Earlier we proposed that the reaction 2NO2 + F2 6 2NO2Fundergos a 2 step
mechanism:
Step 1: NO2 + F2 6 NO2F + F
Step 2: F + NO2 6 NO2F
If the first reaction is the rate determining step, what should the experimental
rate law be for the overall reaction?
If the second reaction is the rate determining step, what should the experimental
rate law be for the overall reaction?
What we have been doing in bits and pieces is using experimental data to try to
prove or disprove a proposed theoretical reaction mechanism. I would like to
formalize what we have been doing
Key Concept
To show that a reaction mechanism is consistent with experimental data you
must show two things:
1. The sum of the elementary steps equal the overall balanced reaction
2. The mechanism revealed in the elementary steps must agree with the
experimental rate law.
Does this prove that the mechanism is correct? NO it only shows it is consistent
with the data. It is usually possible to come up with several different
mechanisms, all of which are consistent with the data. One almost never ‘proves’
a reaction mechanism to be correct
18-5 Reversible Reactions
Up to this point, we have discussed reaction going in 1 direction only, however in
many reactions both forward are reverse reactions are significant (That was one
reason for doing the method of initial rates)
Let’s look at the reaction
2NO(g) + O2(g) 6 2NO2(g)
One proposed mechanism is:
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First, does the sum of the elementary steps = the stoichiometric equation? Yes,
so mechanism might work
Second does the mechanism agree with experimental kinetics?
With the second step being slow you say,’No the problem, I just learned that the
slow step is the rate determining step’ so
rate = k2[N2O2][O2]
The problem is that you have used an intermediate N2O2 in your kinetic analysis ,
and I have said several times that intermediates are very difficult to find, much
less measure their concentration, so we can’t confirm that step with experimental
data!
However, let’s see what the fast equilibrium can do for us. To be at equilibrium
things cannot be changing. That means that the rate of a forward reaction has to
be equal to the rate of the reverse reaction. In this case
the rate of the forward reaction = k1[NO]2
The rate of the reverse reaction = k-1[N2O2]
and
k1[NO]2=k-1[N2O2]
Ah hah I see a cute trick
[N2O2] = k1/k-1[NO]2
and we can substitute this into the previous rate equation:
rate = k2[N2O2][O2]
=k2 @k1/k-1[NO]2@[O2]
=k’[NO]2[O2]
where k’=k2 @k1/k-1
This is indeed the form of the rate law we find experimentally, so this is a
plausible reaction mechanism
Note one could also propose a 1 step mechanism for this reaction in which 1
molecule of O2 and two molecules of NO collide in a termolecular step
Step 1: 2NO + O2 62NO2
rate = k[NO]2[O2]
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This mechanism is also consistent with the experimental data. This illustrates
what I said earlier about you can often find alternate reaction mechanisms that
are consistent with the experimental data.
In this case, most chemists do not like the termolecular step mechanism Can
you think of why? Here is a hint. How often do you see a 3 car crash where all
three cars hit the same place at exactly the same time?
18-6 Catalysis
Key Concept
A catalyst is a substance that increases the rate of a reaction, but is not
consumed in a reaction
How does it do this? It provides a different and faster reaction mechanism that is
available when the catalyst is not present. How can the alternate pathway be
faster? It must have a lower activation energy.
Take example of hydrogenation fo a C=C double bond. When done in the gas
phase you have to have a very high pressure of both gases to get them to
collide. Then the collision has to be oriented correctly so the hydrogens line up
with the double bond, and the collision has to be of a high enough energy to
make the double bond break. When done with a solid metal catalyst (Shown in
power point slide) The hydrogen first gets absorbed onto the metal surface and
the metal reacts with the hydrogen to replace the H-H single bond with metal
hydride bonds to the metal surface. Next the C=C double bond adheres to the
surface. And while it is tethered to the surface it picks up the hydrogens and gets
converted to it product. A totally different mechanism, but this one takes lots
lower pressures of H and C=C double bond and much lower temperatures
because the activation energy is so much lower
You should notice that a catalyst does not change the overall ÄH rxn (as shown in
figure 18.7) so it cannot make unfavorable reaction favorable. All it can do is
speed up a reaction by lowering the activation energy of the reaction.
The example I just gave you is called a heterogenous catalyst because the
solid is a different phase than the gas phase of the reaction. You book gives you
two examples of different homogeneous catalysts that are in the same phase
as the reactants
Key Definitions
Homogeneous catalyst - is in the same phase as the reactants
Heterogeneous catalyst - In a different phase than the reactants
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18-7 Enzyme Kinetics
Your body uses thousands of different enzymes to run all of the biological
reaction needed for life. Most biological enzymes are proteins and we call these
protein catalysts enzymes
Key definition:
An Enzyme is a biological catalyst made from protein.
The reactant in a enzymes catalyzed reaction is called the substrate
The place where the substrate reacts on the enzyme is called the active site.
An example of a biological reaction that is catalyzed by an enzyme is the
reaction of glucose with ATP to form glucose-6-phosphate and ADP
REACTION page 668 This is the first step in the metabolism of glucose to CO2
and a very important reaction for virtually all aerobic organisms
Figure 18.10 gives you an idea of the size of the enzyme compared to the size of
the small molecules it reacts Molar mass of enzymes 10,000 to 50,000 to
millions!
Experimental studies have shown that many enzymatic reactions have
follow the kinetic shown in the next equation:
rate of reaction = ÄP/Ät = k[S]/([S] + KM)
Where [S] refers to the substrate (reactant) concentration
Where k is a constant that is ties to the maximum rate of the reaction
and Km is a constant that ties to how well the substrate binds to the
enzyme
This hyperbolic function looks like figure 18.11
Look at the equation and figure 18.11
at low concentrations of S (left hand side of figure)
S < Km so it drops out of the denominator
and rate ~ K/KM [S] so rate goes up linearly with [S]
At high concentrations of S (right side of curve)
[S]>Km, so Km drops out of the denominator
and rate = k[S]/[S] and the rate hits a maximum rate of k
The book goes over some of the details of a mechanism that has been proposed
to explain the observed kinetics, but I think I will leave that for a Biochem class