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```CHAPTER 14
Chemical Kinetics
Chemical Reactions
There are two things that we are interested in concerning
chemical reactions:
1) Where is the system going (chemical equilibrium, covered in
chapter 15).
2) How long will it take for the system to get to where it is going
(chemical kinetics, covered in chapter 14).
Kinetics is the study of the rate at which a chemical reaction takes
place, and the mechanism by which reactants are converted into
products.
Example
Consider the following irreversible chemical reaction:
A(g)  B(g)
Example (continued)
We may follow the rate of the reaction by either observing the
disappearance of reactant A or the appearance of product B. The average
rate of the reaction over some time period from t1 to t2 is:
Ave. rate =
=
- ( [A]2 - [A]1 ) = - [A]
t2 - t1
t
( [B]2 - [B]1 ) = [B]
t2 - t1
t
where [A] = [A]2 - [A]1
[B] = [B]2 - [B]1
t = t2 - t1
We insert a negative sign when find the rate of disappearance of a
reactant to make the rate of reaction positive.
Example (continued)
Average rate between 10 s and 20 s is (looking at disappearance of A)
Ave. rate = - (0.022 M - 0.030 M) = 0.00080 mol/L.s
(20. s - 10. s)
Instantaneous Rate of Reaction
The instantaneous rate of reaction at time t is equal to the value
for the slope of the tangent line in a plot of concentration vs time
(negative if a reactant, positive if a product).
Stoichiometry and Reaction Rate
Different reactants (and products) may disappear (or appear) at
different rates depending on the stoichiometry of the reaction. For
example:
H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)
In the above reaction ICl disappears twice as fast as H2, and HCl appears
twice as fast as I2. To take this into account we can define the average
rate of reaction as the change in concentration of a reactant or product
divided by the stoichiometric coefficient used to balance the reaction.
Ave. rate = - [H2] = - 1 [ICl] = 1 [HCl] = [I2]
t
2 t
2 t
t
Note we still insert a negative sign when looking at the disappearance of
a reactant. By this method we will obtain the same value for average rate
no matter which reactant or product we observe.
Rate Law
A rate law is an expression that gives the rate of a chemical
reaction in terms of the concentrations of the reactants (and occasionally
other concentrations as well).
For example, consider the following general chemical reaction
a A + b B  “products”
where a and b are stoichiometric coefficients, and we have assumed the
reaction is irreversible (proceeds only in the forward direction).
The rate law for reactions of this type can often be written as
rate = k [A]m [B]n
m = order of reaction with respect to A
n = order of reaction with respect to B
m + n = overall reaction order
k = rate constant, value depends only on temperature. Units
determined by dimensional analysis.
For example, for the reaction
2 NO2(g) + F2(g)  2 NO2F(g)
rate = - [F2] = k [NO2] [F2]
t
Reaction is 1st order in NO2, 1st order in F2, and 2nd order overall.
Note the following:
1) The reaction orders are usually small whole numbers (0, 1, or
2; occasionally 1/2 or -1; rarely any other values).
2) There is no general relationship between the reaction orders
and the stoichiometric coefficients for the reaction. For example, in the
above reaction the stoichiometric coefficients for NO2 and F2 are 2 and 1,
but the reaction orders (determined by experiment) are 1 and 1.
Experimental Determination of the Rate Law
There are several methods that have been developed for finding
the rate law for a chemical reaction. The easiest and most common
method used is the initial rates method.
Consider a general reaction of the form
a A + b B  “products”
where we assume the rate law for the reaction has the form
rate = k [A]m [B]n
For a particular set of initial concentrations, the initial rate of
reaction is the rate of the reaction measured before the initial
concentrations of reactants have had a chance to change significantly.
When we measure the initial rate of reaction, we can then use the initial
concentrations of our reactants in the rate law.
Consider the following two experiments carried out at the same
temperature.
trial 1, initial concentrations of A and B are [A]1 and [B]1, initial
rate of reaction is R1.
trial 2, initial concentrations of A and B are [A]2 and [B]2, initial
rate of reaction is R2.
R1 = k [A]1m [B]1n
R2 = k [A]2m [B]2n
Then
R2 = k [A]2m [B]2n = ( [A]2/[A]1 )m ( [B]2/[B]1 )n
R1
k [A]1m [B]1n
The above is true in general. Now, consider the case where the
initial concentration of B is the same in both trials.
If [B]2 = [B]1, then
( [B]2/[B]1 )n = 1n = 1
and
R2 = ( [A]2/[A]1 )m
R1
We can usually find the value for m by inspection, but for a
rigorous method for determining m we can simply take the natural
logarithm of both sides of the above equation
ln (R2/R1) = ln ( [A]2/[A]1 )m = m ln ([A]2/[A]1 )
m = ln (R2/R1)
ln ([A]2/[A]1 )
The value for n (order of reaction with respect to B) can be found
by comparing trials where the initial concentration of A is the same in
both trials.
Example
The following data were obtained for the reaction
A + B  “products”
trial
initial A
initial B
initial rate
(mol/L)
(mol/L)
(mol/L.s)
1
0.100
0.100
3.1 x 10-5
2
0.100
0.200
3.0 x 10-5
3
0.200
0.200
1.2 x 10-4
We may assume the rate of reaction is given by the expression
rate = k [A]m [B]n
Based on the above data, find m, n, and k.
Finding m. Compare trial 3 and trial 2.
R3 = k [A]3m [B]3n = ( [A]3/[A]2 )m ([B]3/[B]2 )n
R2
k [A]2m [B]2n
1.2 x 10-4 = (0.200/0.100)m (0.200/0.200)n
3.0 x 10-5
4.0 = 2m , and so m = 2
Finding n. Compare trial 2 and trial 1.
R2 = k [A]2m [B]2n = ( [A]2/[A]1 )m ([B]2/[B]1 )n
R1
k [A]1m [B]1n
3.0 x 10-5 = (0.100/0.100)m (0.200/0.100)n
3.1 x 10-5
0.97 = 2n , and so n = 0
So the rate law is: R = k [A]2
To find k we can now use any of the trials. If we use trial 1, then
R1 = k [A]12
k = R1/[A]12 = (3.1 x 10-5 mol/L.s) = 3.1 x 10-3 L/mol.s
(0.100 mol/L)2
Notice that the units for k are determined by dimensional
analysis.
For a real set of experimental data we would find a value for k
from each data set, and then average to find the best value for k.
Also note for real data the values for the reaction orders would
likely not work out to be exactly integers due to experimental error.
For example:
Experimental value
Reaction order
1.94 = 2n
n=1
9.4 = 3m
m=2
1.45 = 2p
p = 1/2
We have assumed in the above that the true values for the
reaction orders should be integer or half-integer values, and that the
small differences we observe are due to random error in the experimental
data.
Typical Types of Rate Laws
As previously discussed, the rate law for a reaction can often be
written as
rate = - [A] = k [A]m [B]n
t
Common rate laws.
Zero order
rate = k
First order homogeneous
rate = k [A]
Second order homogeneous
rate = k [A]2
Second order heterogeneous
rate = k [A] [B]
First Order Homogeneous Rate Law
For a first order homogeneous rate law
rate = k [A]
We may show that concentration vs time is given by the expression
[A]t = [A]0 e-kt
where [A]t = concentration of A at time t
[A]0 = concentration of A at t = 0
k = rate constant (units of 1/time)
A plot of concentration vs time
will exhibit what is called an exponential
decay in the concentration of A.
Finding the Rate Constant
It is difficult to find the value for k (rate constant) from a plot of
concentration vs time since we get nonlinear behavior. We can find a
linear relationship as follows
[A]t = [A]0 e-kt
Take the ln of both sides
ln [A]t = ln [A]0 - kt
(y)
=
b
+ m(x)
y = ln[A]t
x=t
This predicts that for a first order reaction a plot of ln [A]t vs time will
give a straight line, with slope = -k (and intercept = ln [A]0)
One way to test whether or not a reaction is first order is to plot
the logarithm of concentration vs time. If you get a linear result in the
plot, then you know the reaction is first order. If you do not get a linear
result in the plot, then you know the reaction is not first order.
Half-life
By definition, the half-life for a chemical reaction, t1/2, is the time
it takes for the concentration of a reactant to decrease to 1/2 of its initial
value.
For a first order reaction
[A]t = [A]0 e-kt
At t = t1/2, [A]t = [A]0/2,
so [A]0/2 = [A]0 e-kt½
Divide both sides by [A]0
then 1/2 = e-kt½
Take the ln of both sides
Divide by -k
then ln(1/2) = -kt1/2
then t1/2 = - ln(1/2)
k
so t1/2 = ln(2)  0.693
k
k
But - ln(1/2) = ln(2),
Note the following:
1) Since t1/2 = ln(2)/k for a first order reaction, the value for the
half-life is independent of the initial concentration of the reactant.
2) First order reactions are the only reactions where the half-life
is independent of concentration.
3) If we wait two half-lives, the concentration will decrease to 1/4
of the initial value. For three half-lives the concentration will decrease to
1/ of the initial value, and so forth.
8
Second Order Homogeneous Rate Law
For a second order homogeneous rate law
rate = k [A]2
we may show that concentration vs time is given by the expression
[A]t =
[A]0
(1 + kt[A]0)
where [A]t = concentration of A at time t
[A]0 = concentration of A at t = 0
k = rate constant (units of
1/(concentration)(time))
Concentration will not decrease as
quickly in a second order reaction as it
does for a first order reaction.
Finding the Rate Constant
Since [A]t =
[A]0
(1 + kt[A]0)
then if we invert both sides of this equation we get
1 = (1 + kt[A]0) = 1 + kt
[A]t
[A]0
[A]0
(y) =
b
+ m(x)
y = 1/[A]t
x=t
This predicts that for a second order reaction a plot of 1/[A]t vs time will
give a straight line, with slope = k (and intercept = 1/[A]0).
One way to test whether or not a reaction is second order is to
plot 1/concentration vs time. If you get a linear result in the plot, then
you know the reaction is second order. If you do not get a linear result in
the plot, you know the reaction is not second order.
Half-life
We may use the definition of half-life to find an expression for
t1/2 for a second order reaction. It is easiest to do this starting with the
expression
1 = 1 + kt
[A]t
[A]0
If we substitute [A]t = [A]0/2 at t = t1/2, we get (after some algebra)
t1/2 = 1
k [A]0
Unlike a first order reaction, the half-life for a second order
reaction depends on concentration. As the initial concentration decreases
the half life becomes longer.
Zero Order Rate Law
For a zero order rate law
rate = k [A]0 = k
we may show that concentration vs time is given by the expression
[A]t = [A]0 - kt , t < [A]0/k
= 0 , t  [A]0/k
where [A]t = concentration of A at time t
[A]0 = concentration of A at t = 0
k = rate constant (units of
concentration/time)
Concentration decreases at a
constant rate until all of the reactant has
disappeared.
Finding the Rate Constant and Half-Life
Since [A]t = [A]0 - kt , t < [A]0/k
a plot of concentration vs time will be a straight line, with slope = -k,
(and intercept = [A]0).
The half life for a zero order reaction will be
t1/2 = [A]0/2k
Summary of Results
First order
[A]t = [A]0 e-kt
Second order [A]t = [A]0
(1 + kt[A]0)
Sample Problem
Consider the reaction
A  “products”
The reaction obeys first order homogeneous kinetics. The initial
concentration of A in the system is [A]0 0.418 M. After 100. s the
concentration of A is 0.322 M. Find k (the rate constant), and the
concentration of A after 500. s.
Consider the reaction
A  “products”
The reaction obeys first order homogeneous kinetics. The initial
concentration of A in the system is [A]0 0.418 M. After 100. s the
concentration of A is 0.322 M. Find k (the rate constant), and the
concentration of A after 500. s.
[A]t = [A]0 e-kt
ln([A]t) = ln([A]0) - kt
ln([A]t/[A]0) = - kt
k = - (1/t) ln([A]t/[A]0) = + (1/t) ln([A]0/[A]t)
So
k = (1/100. s) ln(.418/.322) = 2.61 x 10-3 s-1.
At 500. s, [A]500 = (0.418 M) exp[-(2.61 x 10-3 s-1)(500. s)]
= 0.1134 M
Temperature Dependence of the Rate Constant
For a reaction that obeys the rate law
rate = k [A]m [B]n
the rate, and therefore the rate constant, for the reaction usually increases
as temperature increases. Experimentally it is found that the temperature
dependence of the rate constant for the reaction often follows a simple
expression called the Arrhenius equation:
k = A e-Ea/RT
where k - the rate constant for the reaction
A - pre-exponential factor
Ea - activation energy for the reaction
If we take the ln of both sides of the Arrhenius equation, we get
ln(k) = ln(A) - (Ea/R)(1/T)
Example:
2 HI(g)  H2(g) + I2(g)
If we assume the data fit the Arrhenius equation, then
ln k = ln A - (Ea/R) (1/T)
and so for each experimental value for k and T we must
1) Convert T into units of K, then find 1/T
2) Find the value for ln k and then plot the results.
The value of Ea is found from the slope of the above plot. The value for
A is then found by substituting one of the data points into the Arrhenius
equation.
x
y
0.0013 K-1
- 4.0
0.0018 K-1
- 15.0
At T = 556. K we have k = 3.52 x 10-7 L/mol.s
x
y
0.0013 K-1
- 4.0
0.0018 K-1
- 15.0
slope = y = [ (- 15.0) - (-4.0) ]
= - 22000. K
x
[0.0018 - 0.0013]K-1
So Ea = - R (slope) = - (8.314 x 10-3 kJ/mol.K) (- 22000. K)
= 182.9 kJ/mol
Since k = A e-Ea/RT, then
A = k e+Ea/RT
At T = 556. K we have k = 3.52 x 10-7 L/mol.s, and so
A = (3.52 x 10-7 L/mol.s) e(182900. J/mol)/(8.314 J/mol.K)(556.K)
= 5.37 x 1010 L/mol.s
Other Forms of the Arrhenius Equation
Beginning with the Arrhenius equation
k = A e-Ea/RT
we can derive the following equation
ln (k2/k1) = - (Ea/R) 1
1
T2 T1
In this equation k1 is the rate constant at T1 and k2 is the rate
constant at T2. By knowing the value for the rate constant at two
different temperatures we may use this equation to find Ea, and then use
the value of k at either temperature to find A.
Collision Theory
The theoretical model on which the Arrhenius equation is based
is called collision theory. The theory makes the following assumptions:
1) For a reaction to take place a collision between reactant
molecules must occur.
2) The reactants must collide with sufficient kinetic energy to
overcome the energy barrier separating reactants and products.
3) The reactants must have a favorable orientation for reaction to
occur.
For example, consider the reaction AB + C  A + BC
Transition state - The species that forms as reactants are
converted into products. It has a structure intermediate between that of
the reactants and that of the products of the reaction.
How will the energy of the system change as we proceed from
reactants to products? We can represent this by an energy diagram.
In this diagram Ea, the activation energy, is the height of the barrier
separating the reactants (AB + C) and products (A + BC). E represents
the change in energy in going from reactants to products. This reaction
is exothermic (E < 0). A---B---C is the transition state. Note that it is a
good approximation to say H  E.
Collision Theory and the Arrhenius Equation
How does collision theory relate to the Arrhenius equation? We
can use the theory to give a physical interpretation to the constants that
appear in the equation.
k = A e-Ea/RT = pz e-Ea/RT
A depends on the collision frequency (z, the number of collisions
between reactant molecules per unit time) multiplied by an orientation
factor (p, the fraction of collisions that have the correct orientation of
reactant molecules).
e-Ea/RT represents the fraction of collisions that have sufficient
kinetic energy to pass over the barrier separating reactants from products.
Notice that as T becomes larger, this term also becomes larger, which
makes sense, as the molecules move faster and have higher kinetic
energy at high temperature than they do at low temperature.
Note: As previously discussed in CHM 1045, vave = (3RT/M)1/2
Orientation
The pre-exponential factor A in the Arrhenius equation depends
on two factors - the rate at which collisions take place and the fraction of
collisions that have a favorable orientation for a reaction to occur.
k = A e-Ea/RT
Example:
Cl(g) + NOCl(g)  Cl2(g) + NO(g)
Favorable orientation
Unfavorable orientation
Reaction Mechanism
A reaction mechanism is a sequence of elementary reactions that
take place on a molecular level and lead from reactants to products.
For example, consider the stoichiometric reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
One possible mechanism for this reaction is
step 1 NO2 + NO2  NO3 + NO
step 2 NO3 + CO  NO2 + CO2
In this mechanism NO3 is a reaction intermediate, a substance
that is neither a reactant nor a product, but which is produced and
consumed as reactants are converted into products.
Types of Elementary Reactions
There are three common types of elementary reactions that
appear in reaction mechanisms.
Unimolecular
A  “products”
rate = k [A]
Bimolecular
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]
Termolecular
Note that the rate for a particular elementary reaction is a rate constant
multiplied by the concentrations of the reactants.
Requirements For a Reaction Mechanism
An acceptable
requirements.
reaction
mechanism
must
satisfy
two
1) The individual elementary steps in the mechanism must add up
to the overall reaction.
2) The rate law predicted by the mechanism must agree with the
experimentally determined rate law.
If the above two requirements are not met, then the reaction
mechanism is not acceptable.
Unfortunately, the reverse is not true. In principle all we can do
with a mechanism is show it is consistent with experiment.
Also note that the final expression for the rate law should not
involve any reaction intermediates.
Finding the Rate Law From the Reaction Mechanism
In general, it is difficult to obtain the rate law for a reaction from
the reaction mechanism (in fact, for complicated systems one often uses
a computer to model the reaction).
There are a few simple cases where we can obtain a rate law from
a mechanism:
1) One step mechanism.
Examples:
O(g) + HBr(g)  OH(g) + Br(g)
rate = k [O] [HBr]
H+(aq) + OH-(aq)  H2O()
rate = k [H+] [OH-]
2) Mechanisms with a single slow step. For a multistep
mechanism with one slow step, the overall rate of the reaction will be the
rate of the slow step. This makes it possible to obtain a rate law as long
as there are no reaction intermediates involved in the slow step.
Example:
step 1 H2(g) + ICl(g)  HI(g) + HCl(g)
slow
step 2 HI(g) + ICl(g)  I2(g) + HCl(g)
fast
overall H2(g) + 2 ICl  I2(g) + 2 HCl(g)
(HI is an intermediate)
Since the overall rate of reaction is equal to the rate of the slow
step, we may say
rate = k1 [H2] [ICl] 1st order in H2, 1st order in ICl
2nd order overall
We can often use experimental data to decide whether a particular
reaction mechanism is possible or not.
Example: Consider the following two mechanisms for the reaction
NO2(g) + CO(g)  NO(g) + CO2(g)
One step mechanism
NO2(g) + CO(g)  NO(g) + CO2(g)
predicted rate law: rate = k1 [NO2][CO]
Two step mechanism
step 1 NO2(g) + NO2(g)  NO3( g)+ NO(g)
slow
step 2 NO3(g) + CO(g)  NO2(g) + CO2(g)
fast
predicted rate law: rate = k1 [NO2]2
Experimentally it is found that this reaction is 0th order in CO and 2nd
order in NO2. That means the first mechanism is not correct, and that the
second mechanism is consistent with the observed rate law.
Fast and Reversible Elementary Reactions
In some reaction mechanisms there will be a step that is both fast
and which goes in both directions.
k1
A+B
C+D
fast, reversible
k-1
Since the reaction is fast in both directions we can assume that
equilibrium is rapidly achieved. At that point, the rate of the forward and
reverse reactions must be the same, so
k1 [A] [B] = k-1 [C] [D]
We can solve the above expression for the concentration of one
reactant in terms of other reactants and the rate constants. This is often
useful in finding the rate law from a mechanism
Example
Consider the following three step mechanism for the stoichiometric reaction
2 H2 + 2 NO  2 H2O + N2
overall reaction
step 1
2 NO
N2O2
fast, reversible
step 2
H2 + N2O2  H2O + N2O
slow
step 3
N2O + H2  N2 + H2O
fast
(N2O2, N2O are intermediates)
What is the rate law predicted for the above mechanism?
step 1
2 NO
N2O2
fast, reversible
step 2
H2 + N2O2  H2O + N2O
slow
step 3
N2O + H2  N2 + H2O
fast
Overall rate = rate of slow step = k2 [H2][N2O2]
N2O2 is a reaction intermediate and so should not appear in the
final rate expression. But from step 1 we may say
k1 [NO]2 = k-1 [N2O2] ; so [N2O2] = (k1/k-1) [NO]2
So rate = k2 [H2] {(k1/k-1) [NO]2 } = (k1k2/k-1) [H2] [NO]2
= koverall [H2] [NO]2
Catalyst
A catalyst is a substance that changes the rate of a chemical
reaction without itself being produced or consumed by the reaction.
Example:
The reaction
2 KClO3(s)  2 KCl(s) + 3 O2(g)
is slow even at high temperatures. However, if a small amount of solid
MnO2 is added to the reaction mixture the reaction will become fast. The
MnO2 is not consumed in the reaction and can be recovered at the end of
the reaction.
Homogeneous catalyst - Appears in the same phase as the
reactants.
Heterogeneous catalyst - Appears in a different phase than the
reactants.
Many catalysts work by lowering the activation energy for a
reaction.
Catalysts can also affect the rate of reaction by providing new
pathways for converting reactants to products, or by changing the value
for the pre-exponential factor (A), usually by forcing the reactants into a
favorable orientation. Biological catalysts (enzymes) often work in this
way.
Example: Hydrogenation of ethene.
C2H4(g) + H2(g)  C2H6(g)
Enzyme - A biological catalyst. Most biological enzymes are
proteins, composed of chains of amino acids, that take on a particular
shape, making it possible to catalyze a specific biological reaction.
End of Chapter 14
“He (van't Hoff) was a modest and unassuming man who never
made priority claims; indeed, in his later books he sometimes gave credit
to others for things - like the ‘Arrhenius equation’ and the ‘Le Chatlier
principle’ - that he had first discovered himself.” - K. J. Laidler, The
World of Physical Chemistry
“Nothing has such power to broaden the mind as the ability to
investigate systematically and truly all that comes under your observation in life.” - Marcus Aurelius
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