Chemical Dynamics
Up until now we have considered thermodynamics which tells us whether a reaction is
spontaneous or not and how much energy is involved in various different forms. Thus, in
principle, we can state whether or not some particular process will happen, but we have
no way of knowing how long that process will take or what path it will take to go from
the initial to the final state. In order to understand this, we must study chemical
dynamics.
The book covers what I consider to be the most important principles in chapters
25 and 27. However, to some extent this material builds upon material in previous
chapters that you have not had. Thus, I will ask you to read chapters 25 and 27, but to use
the material written below as an idication of the subjects that will be covered and the
level of detail that you should use in your studying.
Much of the field of chemistry is discussed in terms of systems which are static at
the time of observation. We perform chemical reactions and consider the structure or
chemical properties of the reactant and product. Yet the chemistry is in the reaction itself,
and that is clearly a dynamic process. One can infer a great deal from the static properties
of molecules, but much of our knowledge of reaction mechanism relies on the
observation of the time dependence of chemical reactions.
What can one learn from reaction kinetics? Most of the time in studies of
chemical kinetics, what one is doing is relating the time evolution of the population of
some molecular species (the observable) to a model for the mechanism of the reaction
taking place. Several types of information can come out of such a study. First, one can
learn what the path of the chemical reaction is. This information typically comes from
comparing the expected results from a series of possible kinetic pathways to the actual
time dependence of the molecular population observed. Next, one can learn what the
specific rates are for different reaction steps. This in turn provides information about
what limits the rate (and often the yield) of chemical reactions. By varying conditions,
such as temperature or solvent type, one can learn what type of energetic barriers must be
overcome in order for a reaction to proceed.
A few applications of chemical kinetics. Kinetic analysis has been applied to
many systems in order to learn more about the reaction mechanism. For example, let's
say that you are converting one mixture of chemical species, A and B, to another mixture,
C and D, by heating for a period of time and you want to learn something about the nature
of the reaction. One simple question you can ask is, "what type of reaction limits the rate
of the overall conversion?". By looking at the kinetics of the reaction as a function of
reactant concentrations, one can determine if the rate limiting step is the formation of an
interaction complex between the reactants, the resolution of this complex into the
products, or a unimolecular conversion of one of the reactants or reaction intermediates to
a different form before the reaction can take place.
One common application of kinetics in biochemistry is the study of enzyme
kinetics. An enzyme is a catalyst, and the kinetics of enzymes is similar to the kinetics of
other catalytic reactions. Questions that can be addressed by looking at the kinetics of an
enzyme reaction are: how strong is the interaction between the reactants or products and
the enzyme, are there other molecules which affect the rate of the reaction (but are not
themselves used up) besides the enzyme, and are reaction rates limited by diffusion?
Another interesting application of standard kinetic theory is electron transfer
reactions in a solid state (i.e. first order) system. This is often used to model electron
transfer in either respiratory or photosynthetic systems in biochemistry. In this case, one
typically looks at either the optical spectrum or the electron paramagnetic resonance
spectrum as a function of time to identify reaction intermediates and to investigate the
reaction mechanism.
During the course of the next two weeks, we will consider the basis for multiorder
and first order kinetics, with emphasis on the application of kinetic information in
2
determining reaction mechanism. Then we will discuss in more detail what actually
determines the individual rate constants of a reaction.
The rate of a chemical reaction. Chapter 25.2. Much of the experimental aspect
of chemical kinetics is concerned with the measurement of reaction rate. There are a
couple of points worth making about this. First, the rate of change in the concentration or
population of some reaction component with time is given by
d  A
rate 
dt
This is not always as straightforward as it sounds. First, how you define the rate of a
reaction may well depend on which reactant or product you are interested in. For
example in the reaction A B the rate of the reaction could either be expressed in terms
of the change in [A] with time or the change in [B] with respect to time. The two rates
are just negative of one another:
d  A
d B 

at
dt
To overcome this, we introduce the general expression for a reaction rate
1 d J 
v
 j dt
where j is the stochiometric coefficient in front of component J of the reaction and j is
negative for reactants and positive for products. Thus for the reaction above, the reaction
rate is just
dA dB
 

dt dt
For a more complicated reaction like 2A + B  C
1 dA
dB dC
v


2 dt
dt
dt
3
Second, rates are typically concentration, and therefore time, dependent. Thus, there is no
single rate for a reaction. To simply say, reaction A is faster than reaction B has very
little meaning without specifying all the concentrations of the components involved. For
example, consider the rate at which marbles come out of a small hole in a container. If
the container has many marbles in it, then many marbles happen to hit the hole and pop
out in any given period of time. However, if the number of marbles is small, then the rate
at which marble will come out is slower.
As chemists, we would like to be able to compare reaction kinetics quantitatively. Thus,
we need to come up with some parameters related to the reaction kinetics which are
concentration independent.
Rate constants. Consider the simple irreversible (i.e. no back reactions) reaction
A B
4
The instantaneous rate of the reaction at any time, t, is just the derivative
dAt 
. Note
dt
that for simplicity I will sometimes refer to the time dependent concentration of a
molecule as, for example, A(t), rather than using brackets, [A]. Now, what will this
reaction rate depend on and what should the functional form be? What assumption is
used to relate the rate of A's demise to its instantaneous concentration in a linear manner?
The fundamental assumption is that individual molecules have no memory. In other
words, all A molecules have the same probability of converting to B in a given period of
time regardless of their history. More generally, we assume that any particular molecule
of A is identical to any other not only at the present time but at any past or future time. If
this is true, than the instantaneous probability that any particular molecule of A will
change to B is time independent. This instantaneous probability, then, is a useful
constant for comparing irreversible first order reactions.
Ok, but what is an instantaneous probability and how does it relate to a normal
probability? Well, let's say you are star gazing out in the woods at night and you are
looking for shooting stars (OK, meteorites, since this is a science class). You watch for
30 minutes and you see 33 meteorites. What is the probability of seeing a meteorite in
one minute? Well, we can see that the average number of meteorites per minute is 1.1
(33/30). Clearly, however, the probability of a meteorite coming each minute could not
be 1.1. What we have ignored is the fact that sometimes there will be two meteorites in a
minute and sometimes 3, etc. To get the number of shooting stars we see per second, we
need to add up all of these probabilities times the number of events:
meteorites
  # meteoritesP# meteorites
sec ond
Where P(#meteorites) is the probability of seeing that number of meteorites in a minute.
Now the probability of seeing two meteorites in a minute is just the probability of seeing
5
one meteorite squared. The probability of seeing three is just the probability of seeing
one to the third power, etc.:
meteorites
 0 P0   1P1  2 P2   3P3  ...
sec ond
2
3
 P1  2 P1  3P1  ...
I suppose we could try and solve this equation given for the probability of seeing one
meteorite in a minute and then use this to calculate the total probability of seeing one or
more meteorites in a minute, but it would certainly be complex. We need to solve the
problem a different way. It would be nice if all the squared terms and cubic terms, etc.
dropped out. This will happen if the probability of the event in the given period of time is
small enough. If we go to a really tiny time period, then it would be the case that:
meteorites
 P1
sec ond
The instantaneous probability is the probability per unit time in the limit that the time
period considered in very, very small. So for the case of the meteorites, the instantaneous
probability is 0.01833 1/s. We can get the actual probability from this for a very short
period of time by simply multiplying by the time. Thus the probability of seeing a
meteorite in 2 seconds (still very short compared to the normal interval between
meteorites) is approximately 0.01833 x 2. The strange thing is that I can express the
instantaneous probability in any units that I like. I could also say that the instantaneous
probability of seeing a meteorite is 1.1 (1/min). However, I would not get the probability
of seeing a meteorite in one minute by multiplying this by 1 min because one minute is
not short compared to the interval between meteorites.
Using the concept of an instantaneous probability, we can now write an equation. The
dAt 
fractional change in the concentration of A,
, is just going to be given by the
At 
instantaneous probability of A converting to B multiplied by the increment of time which
has passed (as long as that increment is short):
6
dAt 
 k  dt
At 
(1)
where k is the instantaneous probability of A converting to B. Rearranging:
dAt 
 k  At 
dt
(2)
and now we can see that the instantaneous probability of A converting to B is identical to
what we chemists have always called a first order rate constant, k. This is important. The
physical meaning of a first order rate constant is that it is the instantaneous probability of
one chemical species converting into another. This has some interesting consequences.
For example the expression k  t is the true probability that A will convert to B in
1
timet as long as t  .
k
Now that we know how to calculate the fractional change in A(t) in any increment
of time, dt (eq. 1) we can simply integrate this equation to determine A(t) resulting in:
dAt 
  k  dt
At 
dAt 
 At     k  dt
ln At    kt  c
At   e  kt  c  e c e  kt
Where c is just the constant of integration. Thus, e c is just a constant. Let's call it A0
and we can see that
At   A0 e  kt (3)
7
What does A0 physically represent? Consider what this equation looks like at time zero.
When t=0, e0=1, so A(t) = A0. In other words, A0 is the initial concentration of A.
Now here is an interesting question -- How did e get into all of this anyway? We have
seen e pop up in thermodynamic equations before. Mathematically, of course, it came
dx
dAt 
from the integral of
(or in this case
, but there is no mathematical need to leave
x
At 
eq. (3) in terms of e. We could write
At   A0 2  k t
'
where
k
k' 
ln 2
(4)
Why not define things this way? 2 is a much easier number to remember than e. The
reason is directly related to the definition of k given above. The first order rate constant
will only have the physically meaningful identity of a true instantaneous probability if the
integrated time dependence expression is phrased in terms of e. This is why in physics e
always pops up in situations were the change in one parameter can be related to the value
of another in terms of a probability density. Another example common in chemistry is
thermal distributions such as K eq  e  G / RT . You can think of the equilibrium constant as
the result of a thermal distribution that defines the probability of being in the initial vs.
the final state. It's really the same kind of problem.
Next lets consider a simple second order reaction:
A B C
8
The way chemists normally think about this, it is usually broken down into two processes
(more on the theory behind this idea later):
A  B  A B  C
where A  B is a collision complex between A and B. The second part is simple. That is
essentially just the first order reaction we talked about above, one complex, A  B,
changes to another with some instantaneous probability, k. However, this probability
must be multiplied by the probability that any particular molecule (lets consider this from
the point of view of the A molecules) will be in the complex form during an interval of
time, dt. What we can do now is to define some interaction volume around B and define
any A molecule that is within that volume to be part of a collision complex with B. We
now have a choice about how we want to think about the reaction. It could be that the
reaction occurs every time that A comes within B's interaction volume or it could be that
A will pass in and out of B's interaction volume many times before the reaction occurs.
This will change the molecular interpretation of the rate constants but not their form. For
simplicity, we will assume now that the collisions are elastic, that is, that A rapidly
bounces off of B and only remains in the interaction volume for a period which is short
compared to the inverse of the first order rate constant of the second part of the reaction.
In this case, the overall spatial distribution of A relative to B will always be random and
we can use the molar interaction volume to determine the number of A molecules which
are interacting with B:
A  B  A x (Fraction of A within B's interaction volume)
Fraction of A within B's interaction volume = B x (B's molar interaction volume)
so, A  B  A x B x (B's molar interaction volume) (5)
9
Note that we express A and B in terms of concentrations (moles/liter) and B's molar
interaction volume in liters per mole, so the units of the right side of the equation are
correct (a simple concentration).
Now using the first order rate expression for formation of product:
dC
 k A  B
(6)
dt
and substituting in eq. 5:
dC
 k  VB  A  B
dt
(7)
Where VB is the molar interaction volume of B. We normally refer to k  VB as the second
order rate constant for the reaction. Note that the specific interpretation of the second
order rate constant as the product of a true instantaneous probability and a molar
interaction volume is dependent on the assumptions that the collisions between A and B
are elastic and not usually productive (many collisions are required before a reaction
occurs). If we relax these assumptions, the rate equation still has the same form, but the
molecular interpretation of the second order rate constant is somewhat different. Also
note that the molar interaction volume of A and B are by definition the same (since this
volume only applies to interactions between A and B and this interaction is symmetric).
Thus, it makes no difference whether we look at the problem from the point of view of
molecule A or from the point of view of molecule B.
The order of a reaction. I have been loosely talking above about first and second order
reactions. I spoke of reaction order as though it could be determined by the number of
chemical species that had to come together in order for the reaction to occur. In fact,
reaction order is really an empirical quantity which does not even need to be an integer. It
is simply given by the sum of all the exponents on the concentration terms in equations
like
dA
2
 k  AB 
dt
10
if one finds that in fact emprically this equation is true (regardless of what the chemical
reaction equation looks like) this would be called a third order reaction (the exponents are
1 + 2). I am not really much interested in empirical reaction orders here. I would rather
we worked with simple cases that correspond directly to the mechanism of the reaction.
The book calls these types of reactions elementary reactions and instead of talking about
reaction order uses the term molecularity. I am going to stick with simple cases where
the reaction order equals the molecularity and thus we can determine the order of the
reaction just by looking at the chemical equation. The easiest way to get used to the
relationship between rate, rate constants, reaction order (molecularity) and rate equations,
let's consider a few examples:
Reaction
AB
A+BC
AB+C
AB
A+BC
2A+B  C
A+B+CD
Rate equation for A
d  A
  k  A
dt
d  A
  k  AB 
dt
d  A
  k  A
dt
d  A
  k f  A  k b B 
dt
d  A
 k f  AB   k b C 
dt
d  A
2
 2k  A B 
dt
d  A
 k  AB C 
dt
Units of rate
constant
Reaction order
(molecularity)
s-1
1
M-1s-1
2
s-1
1
s-1, s-1
1,1
M-1s-1,s-1
2,1
M-2,s-1
3
M-2s-1
3
Equilibrium in terms of dynamics. One of the connections between thermodynamics and
kinetics is in terms of the concept of equilibrium, Thermodynamically, equilibrium is the
ratio of products and reactants that results in the lowest Gibbs free energy for the system
11
(or the point where the chemical potentials of the products and reactants are equal).
Kinetically, equilbrium is the point where the rate of the forward reaction is equal to the
rate of the reverse reaction: product is going to reactant at the same rate as reactant is
going to product. At this point, the net change in both reactants and products is zero. For
the reaction A B:
dA dB

0
dt dt
In a situation where we are considering reactions that can go either direction (and all
reactions can at least to some small extent), there is a forward and backwards rate
constant. These can be first or second (or higher) order rate constants, depending on the
chemistry. Also, the order in the two directions can be different. AB+C is first
order in the forward direction but second order in the reverse direction. In the case where
both forward and reverse reactions are first order, AB, if we call the forward rate kf
and the reverse rate kr we can write that
dA
  k f  A  k r B   0
dt
k f  A  k r B 
B   k f
A k r
 K eq
Thus, one can see that the equilibrium constant for a reaction is simply given by the ratio
of the forward to reverse rate constants. As an exercise, show that this is also true for the
reaction A+B C (the equilibrium constant is now the ratio of the second order
forward rate constant to the first order reverse rate constant). You might get a problem
like this one on an upcoming exam.
Reaction Sequences (this follows roughly 25.7).
Many of the more interesting reactions you will come across do not occur as a simple
single reaction, but in fact occur as a series of reactions. Let's consider a reaction
A  B+C  D
12
For this case lets consider the first forward reaction to have a rate constant of kf1, the first
backward reaction to have the rate constant kb1, the second forward rate constant is kf2
and the second backward rate constant is kb2. How do we describe the kinetics of this
reaction? We can always write down the rate equations for each species in the reaction,
assuming that these are elementary reactions which are an accurate representation of the
reaction mechanism (this is something you should be prepared to do):
dA
 k f 1  A  k b1 B C 
dt
dB
 k f 1  A  k b1 B C   k f 2 B C   k b 2 D 
dt
dD
 k f 2 B C   k b 2 D 
dt
By the way, you should also be able to tell me the initial rate of change in the
concentrations of A, B, C and D if you are given the initial concentrations of each and the
values of the rate constants.
But how do we go farther than this to determine the actual concentration time
dependence? The answer is that we can either solve these differential equations directly
(which can be done for this case, but is messy) or we can make some simplifying
assumptions and reduce it down to a simpler case.
For rate equations up to about the complication of the one in the example above, it
is fairly straightforward to solve the equations exactly, especially if one uses programs
such as Maple or MathCad to help with the algebra. More complex series of chemical
reactions often have no analytical solution. In these cases we have two choices. We can
either solve our problem numerically (calculate the change in concentration for a small
change in time, update the concentrations and recalculate for the next time increment,
etc.) or we can use an assumption about the way that the chemical mechanism works.
Here, I would like to go over three such assumptions.
Limiting rate assumption. This is probably the simplest of assumptions. Let's say
I have a series of reactions that are essentially irreversible, A B  C. In other words,
13
the equilibrium constants are much greater than 1. Let's also suppose that the first
reaction is much faster than the second. What will be the time dependence of product
formation? It will be determined simply by the slowest reaction. In this case, the time for
the B to C reaction will be limiting. Thus, the overall rate of the reaction will be
determined simply by the rate constant of the B to C reaction. Note that the rate of A
disappearance will be determined by the rate constant of the A to B reaction. This is
because it does not matter to A what happens later in the chain of reactions.
Equilibrium assumption (called pre-equilibrium in the book). Here, we assume
that one of the reactions comes to equilibrium very rapidly compared to the other rates in
the reaction. Consider the reaction A  B  C. I am not going to prove this to you
(try it yourself if you like), but it turns out that the rate constant for approach to
equilibrium in a simple case like this is just the sum of the forward and backward rate
constants of a reaction. Thus we can say in this case that if the sum of the forward and
backwards rate constants for the first reaction is much faster than the rate constant for the
second reaction then the first reaction will come to equilibration before much product is
formed. (Generally, what follows is true any time the first reaction comes to equilibrium
rapidly compared with the time scale of product formation.)
dC
 k 2 B 
dt
B   K
A eq
B   K eq A
dC
 k 2 K eq  A
dt
here k2 is the rate constant for the second reaction and Keq is the equilibrium constant for
the first reaction. We have now reduced this series reaction to something that is
mathematically like a first order reaction of A going to C, except that the effective first
order rate constant is comprised of both k2 and Keq. Let's do a real example here. Let's
say that I add one mole of A to a liter of water. I want to know what the time course of C
14
formation is. Ok, the equilibrium assumption says that instantly A and B will come to
equilibrium. Thus, the amount of A and B present just after adding A in the solution is
given by the equilibrium constant. Let's suppose that the equilibrium constant is 0.01 and
that the rate constant for the second reaction is 1 1/s. Thus the equilibrium concentration
of A is 0.99M and that of B is 0.01M (A and B come to equilibrium instantly giving a
0.01 ratio of B to A). This means that effectively, the initial amount of A is 0.99M and
the effective first order rate constant for formation of C from A is k2Keq = 0.01 1/s. Thus
the time course is given by C(t) = 0.99M e-(0.01)t. Note that since this reduces to a first
order rate equation the integral is the same as it was for a first order rate equation
previously. Note also, that this would not have been so simple if the equilibrium constant
had been closer to 1. Here we were able to largely ignore the contribution of B is the
overall concentration. If the equilibrium constant were closer to one, this would not have
been possible.
Steady-state approximation. The steady state approximation assumes that the
intermediate is present not in equilibrium with one of the other chemical species but
rather in steady-state. This means that it is being used up as fast as it is formed and
therefore its concentration does not change very quickly. In this case, we have that
dB
 0  k f 1 A  k b1 B  k 2 B
dt
 k f 1 A  k b1  k 2 B
thus,
B  k f 1
A k b1  k 2 
k f1
dC
 k2 B  k2
A
k b1  k 2 
dt
Again, we have reduced the series of reactions to something that looks rather like a first
order reaction of A forming C. Again, as long as the intermediate concentration of B is
15
low, we can simply analyze the system as though it was a first order reaction between A
k f1
and C with a first order rate constant of k 2
.
k b1  k 2 
Predicting rates -- an introduction to kinetic theory (this is mostly in chapter
27, but described quite differently). What we have discussed thus far is primarily how
to take a mechanistic model and, given the rates, predict the time dependence or the
concentration dependence of a chemical reaction. This is all really not based on
chemistry, but rather on probability theory and the behavior of populations. The
chemistry is buried in the rate constants that we use in the models. But, how does one
actually predict what the molecular rate constants for a reaction would be? This is not
easy to do in a quantitative sense for most chemical systems, at least not in terms of
deriving an absolute value for the rate. However, we can predict with some reliability the
effect of different environmental factors on the rate constants for the system. One such
effect that you are probably all familiar with is the effect of temperature on rate. You
have probably all seen pictures like the one given below:
16
What is this a picture of? It is a plot of the potential energy of the system as one
progresses along the reaction coordinate connecting the initial and final state. The
potential energy is the Y-axis and the reaction coordinate is the X-axis. Between the
initial and final state is an energy barrier. In order to get over that barrier, energy must be
put into the system from the bath. The availability of that energy is going to be very
temperature dependent. This gives rise to the concept of an activation energy required to
overcome the barrier and the expression:
k  e  E A / RT
(31)
Here EA is the activation energy and is in joules per mole as written. You will
sometimes see -EA/kBT (instead of -EA/RT). Expressed as -EA/kBT it is the activation
energy for one molecule divided by the temperature times the Boltzman constant. The
Boltzman constant is just R/N where N is Avagadro's number. (This expression actually
results from statistical mechanics. It is the probability that any individual molecule or
molecular complex in the system will have an energy greater than EA which is required to
overcome the barrier.) This, then, allows us to relate rates to some parameters of the
energetics of the system, specifically the energetics of positions intermediate between the
initial and final states of the system. But how do we think about the potential energy
picture above on a molecular level and what factors will determine the activation energy
for any particular reaction? One approach to understanding this is transition state theory.
The basic idea is that the activation energy associated with a reaction can be attributed to
the formation of a particular transition state, such as the high point between reactants (R)
and Products (P) above. This could be a particular conformation of a molecule or a
collision complex between molecules. Let's consider a couple of models:
17
In order to understand this, it is important to talk in more detail about what the
reaction coordinate in these potential energy diagrams really means. The reaction
coordinate that one draws in these two-dimensional pictures is some composite nuclear
coordinate that is supposed to represent a change in the nuclear configuration of the
chemical system which accompanies the reaction. In a very simple molecular system,
such as NO, this could represent the distance between two bonded atoms. In more
complex molecular systems, this coordinate may represent distance changes or orientation
changes of many bonds in the system which are required to
get from the reactant to the product.
In transition state theory, one talks about the
progression from reactant to the product as though it occurred
through a distinct intermediate in the system. This
intermediate sits at the highest point in the potential surface
between the reactant and product and the system can proceed in either direction from that
point in the reaction. The usefulness of this concept comes from the fact that now one
can suggest a particular structural model for what that transition state is and then from
this predict how the rates and yields of a reaction will depend on external influences. For
example, electrophillic addition to simple alkenes takes place in such a way as to form the
most stable intermediate carbonium ion. Thus, more polar solvents, which will solvate
and therefore stabilize this ionic intermediate, are likely to speed up such a reaction. This
can be turned around from an experimentalist's point of view. If we postulate that the
intermediate in a reaction is a carbonium ion, than we should see a predictable effect of
solvent polarity on reaction rate. One might also be able to predict the effects of
structural changes to the molecule or the presence of other molecules which might
stabilize or destabilize the proposed transition state. This is a simple and elegant concept
which has been the basis of many if not most of the mechanistic schemes developed for
chemical reactions.
18
How would we predict the activation energies and rate constants of such a second
order reaction? Well, that depends on exactly how the reaction worked. When we first
discussed second order reactions, we talked about one way of determining the rate
constant by considering the interaction volumes around A and B and estimating the
probability of collision. We would then need to know how much energy was imparted by
the collision. That is determined by RT.
Another approach is to look at the system as an equilibrium between the reactants
and the transition state. In this case,
k
A  B  AB 

C
dC
 k  AB
dt
AB
K eq 
AB
dC
 kK eq  AB 
dt
Here again, we can easily see the exponential dependence of rate on temperature,
K eq  e

G
RT
G

dC
 ke RT  AB 
dt
In this case, the activation energy is just the Gibbs free energy between the reactants and
the transition state.
19