A.P. Chapter 13 Outline: Chemical Kinetics
 Chemical Kinetics: the study of the speeds of reactions and the
nanoscale pathways or rearrangements by which atoms and
molecules are transformed from reactants to products.
I.
Reaction Rate
A. Homogeneous reactions: reactions in which reactants and
products are all in the same phase
B. Factors affecting the speed of a homogeneous reaction:
1) the properties (bonding and structure) of reactants and
products
2) the concentration of reactants and sometimes products
3) the temperature at which the reaction occurs
4) the presence of a catalyst (a catalyst speeds up a
reaction but undergoes no chemical change itself)
C. Heterogeneous reactions: reactions that take place at an
interface between two different phases, such as solid and gas.
Their speeds depend on the four factors above as well as the
area and nature of the surface at which they occur.
D. Reaction Rate: the change in concentration of a reactant or
product per unit time.
Rate =- [X]2 – [X]1 = - ∆[X]/∆t
T2 – t1
E. Reaction Rate and Stoichiometry: the rate of change in
concentration of any reactants or products is multiplied by the
stoichiomtric coefficient to find reaction rate.
Example: for the reaction 2 N2O5 (g)  4 NO2(g) + O2(g), if one is
looking for the rate of change of concentration for N2O5, rate = ½ ∆[
N2O5]/∆t
rate of change of concentration for NO2, rate = ¼ ∆[ NO2]/∆t
rate of change of concentration for O2, rate = ∆[ O2]/∆t
F. Average Rate and Instantaneous Rate
 A reaction rate calculated from a change in concentration divided
by a change in time is called the average reaction rate. Generally,
average rates become smaller as the concentrations of one or more
reactants decreases.
 The instantaneous reaction rate is the rate at a particular time after
a reaction has begun. The instantaneous rate is the slope of the line
tangent to the concentration-time curve at the point corresponding
to the specified concentration and time.
II.
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Effect of Concentration of Reaction Rate
The rate of reactions are proportional to concentrations
Rate = k[X] where k = rate constant, which is independent of
concentrations but not temperature.
The relation between rate and concentration (the rate law) must be
determined experimentally
The initial rate of a reaction is the instantaneous rate determined at
the very beginning of the reaction
To approximate initial rate calculate -∆[reactant]/∆t after no more
than 2% of limiting reactant is used up.
III.
Reaction Rate Law
Law of Mass Action: The rate of a chemical reaction is
proportional to the product of the concentrations of the
reactants.
For any general reaction
aA + bB ------->
the rate law expression is: r [A]m[B]n
where A and B represent the molar concentrations of A and B.
m and n are the powers to which the concentrations must be
raised. k is a constant of proportionality known as the rate
constant. Data show that the rate constant is not affected by []
(concentration) changes but does vary with temperature
changes. The values of 'm' and 'n' are not the stoichiometric
numbers obtained from the balanced equation; unless; the
equation is deemed to be a one-step reaction, but more on this
latter. The only valid way to obtain the values of m and n is to
use experimental data.
The exponents, m and n may be zero, fractions or integers.
The sum of the exponents is called the reaction order.
eg. H2(g) + I2(g) -----> 2 HI(g)
r = k[H2][I2]
This is a second order reaction. The sum of the components is 2.
In this case the values of 'm' and 'n' just happen to be the same
as the stoichiometric numbers in the balanced equation.
Therefore it must be a one-step reaction. i.e., there is only the
one reaction step needed to convert reactants into products.
This is not always the case and a simple reaction may proceed
through a number of intermediate steps.
Experimental Determination of Reaction Order
The rate is usually given in terms of moles/Litre seconds but this
is not always the case. You will have to be very careful about the
units of each participant reactant in order to get the proper units
for the rate constant itself.
Before we can determine the value of the rate constant 'k', we
need to find the exponential value for each participant. To find
the relationship of one reactant it is necessary to keep the other
reactant(s) constant so look at the following reaction and data
table:
NO + H2 -----> HNO2
(balance it if you'd like, but the balanced equation will not give
you the exponential values unless the equation is a one-step
reaction).
Rates of reaction between NO and H2 at 800oC
Experiment NO
H2 Initial Rate of Reaction
Number
moles/L moles/L
moles/L sec
1
0.001 0.004
0.002
2
0.002 0.004
0.008
3
0.003 0.004
0.018
4
0.004 0.001
0.008
5
0.004 0.002
0.016
6
0.004 0.003
0.024
----------------------------------------------------------------From the equation we can write a partial rate law as
rate = k[NO]m[H2]n
You have 6 experiments to choose from. Using these 6
experiments you must determine the values of 'm', 'n' and 'k'.
There are only two reactants, so choose one to start to work
with. We'll start with NO. Choose any two experiments where
the concentration of NO changes but the concentration of H2
stays the same. We want to determine how the NO changes the
rate! We will choose experiments 1 and 2.
Using experiments 1 and 2 you can see that the concentration
jumps from 0.001 to 0.002 moles/L. IT DOUBLES!! Take a look
at the rates for these same experiments. The rate jumps from
0.002 to 0.008 moles/L seconds. IT QUADRUPLED!!.
The exponential constant 'm' for the [NO] is the mathematical
relationship between these two values. i.e.
2m = 4 therefore m = 2 because 22 = 4
To confirm this, compare experiments 1 and 3. The
concentration of the NO gas TRIPLES. The rate jumps by a
factor of nine. Therefore
3m = 9 so m = 2 because 32 = 9
The rate law expression can now be updated to: rate =
k[NO]2[H2]n
We will now use experiments where the [NO] concentration is
kept constant and the [H2] changes.
Look at experiments 4 and 5. The H2 concentration DOUBLES
and the rate DOUBLES.
2n = 2 therefore n = 1 since 21 = 2
To confirm this number look at experiments 4 and 6. The H2
concentration TREBLES from 0.001 to 0.003 The rate also
TREBLES from 0.008 to 0.024
3n = 3 therefore n = 1 since 31 = 3
So the rate law expression can be rewritten as
rate = k [NO]2 [H2]1
From the sum of the exponents this is a third order reaction.
Now to determine the value of 'k'. 'k' is a constant. It's value
should not change (except under temperature changes). Choose
any one of the experiments. It should not matter which one.
We will use experiment 1. Using the rate law, above fill in the
values from the data table.
0.002 mol/L sec = k (0.001 mol/L)2 * (0.004 mol/L)
0.002 mol/L sec = k * (0.000001 mol2 /L2) * (0.004 mol/L)
0.002 mol/L sec = k * 0.000 000 009 mol3/L3
k=
0.002 mol/L sec
0.000 000 004 mol3/L3
= 500,000 sec/mol2 L2 or sec mol-2 L-2
Therefore the rate law equation for this reaction is
rate = 500,000 sec mol-2 L-2 [NO mole/L]2 [H2 mol/L]
IV.
Half-Life
 The half-life of a reaction, t1/2, is the time required for the
concentration of a reactant A to fall to ½ of its initial value.
 T1/2 = -ln2/-k = 0.693/k
V.
Temperature and Reaction Rate: The Arrhenius equation
 The most common way to speed up a reaction is to increase the
temperature.
 Rate constants are temperature specific
 K = Ae-Ea/RT where A is the frequency factor, e is the natural
log base, Ea is the activation energy, T is the Kelvin temperature
and R is the gas constant 8.314 j mol-1K-1
Activation Energy and the Activated Complex
This is the energy that must be reached by 2 colliding molecules before a reaction
can take place.
If enough energy is supplied to the boulder to push it over the hill (activation
energy barrier), it will spontaneously roll down the mountain, releasing energy
as it moves to a lower state of potential energy. The rate of boulders being
pushed over the cliff will depend on the height of the activation energy barrier.
Collision Theory
The rate of a reaction depends on two factors.
1) The number of collision per unit time between the reacting species.
2) The fraction of these collisions that are successful in producing a mew
molecule.
Collision Geometry
If two or more molecules collide but are not orientated correctly then no
reaction will take place.
For a reaction to occur, molecules must collide not only with sufficient
energy but with the proper orientation.
Why is there an Activation Energy Barrier?
During the course of a reaction considerable redistribution of electrons may occur.
Consider, for example, the reaction of CH3Br with Cl- in water at 298K.
As the bimolecular reaction occurs (i) there is angle bending: the initially pyramidal
CH3 grouping become planar; (ii) there is bond-making and breaking: a partial CBr
bond is weakened. The energy released by the formation of the partial CCl bond will
not fully compensate for the other two (endothermic) changes and yet there is no lower
energy pathway from reactants to products. The reactants can get to the point of
highest potential energy (the "activated complex" or "transition state" - in curly braces
above) only if they initially have sufficient kinetic energy to turn into the potential
energy of the activated complex. The activated complex can not be isolated; it is that
arrangement of reactants which can proceed to products without further input of
energy.
It is often useful to make a schematic plot of the total energy (enthalpy) of the combined
reactant molecules during the various stages of the chemical reaction. The points on the
plot which we can pinpoint are:
(i) the difference between the average energy of the products and the average energy of
the reactants, Hreaction (~ + 25 kJ mol- for CH3Br + Cl- ) and
(ii) the activation energy (obtained experimentally; 103 kJ mol- for CH3Br + Cl-).
If we assume the total energy varies smoothly with the course of the reaction we obtain
the following "energy profile":
It is important to note that there will be more than one way for the reactants to interact
and so pass to the products. However, there can only be one minimum energy
pathway and essentially all of the reaction will occur via this pathway.
Effect of Catalysts on the Activation Energy
Catalysts provide a new reaction pathway in which a lower A.E. is offered.
A catalyst increases the rate of a reaction by lowering the activation energy
so that more reactant molecules collide with enough energy to surmount the
smaller energy barrier.
Enzymes, Biological Catalysts
enzyme inhibitor
molecule + molecules
enzyme-inhibitor
complex
enzyme blocked
from joining
A&B
An inhibitor molecule with a shape simlar to the enzyme's can block the
ability of the enyzme to catalyze the reaction of other molecules.
Why Do Catalysts Work?
Solid catalysts have large surface areas and are capable of adsorbing the
reactants onto their surfaces. One of the reactants molecules may readily
react with the atoms of the catalyst and form an intermediate species. This
new intermediate species then reacts readily with the second reactant. The
desired product is formed leaving behind the catalyst.
Relationship Between The Activation Energies of Opposing
Reactions
Slow reactions have high activation energies. Fast reactions have relatively
low activation energies. An endothermic reaction always has a greater
activation energy and a slower rate than the opposing exothermic reaction.
An increase in the temperature affects the rate of the endothermic reaction
more than that of the exothermic reaction.
Reaction Mechanisms
An elementary process is a one-step process in which the product particles
are in most cases the direct result of collisions of only 2 reactant particles.
There are three particles collisions but they are rare. The chances of 4 or
more particles forming a one-step reaction reach infinity. The reaction
mechanism for
H2(g) + Cl2(g) ---> 2 HCl(g) is made up of 4 elementary steps.
Rate Determining Step
When a reaction is the result of a series of elementary processes, the rate of
the overall reaction is determined by the slowest reaction in the sequence.
eg. Ag+(aq) + Cl-(aq) ---> AgCl(s)
a simple reaction involving only a single 1 step mechanism therefore very
fast.
eg. 2 NO(g) + 2 H2(g) ----> N2(g) + 2 H2O(g)
This appears to be a 4 particle reaction as written. Impossible!!
The reaction mechanism for the above reaction has been determined to be:
step #1 2 NO + H2 ----> N2 + H2O2 slow since it invovles a rare three particle
collision. This is the slowest and therefore is the step that determines the
rate of all the other steps.
step #2 H2 + H2O2 ------> 2 H2O fast reaction
Deduction of Rate Laws from Mechanisms
The mechanism of a reaction is the series of elementary steps by which the
reaction takes place. We have already seen that in a series of steps there will
usually be one (the slowest or rate-determining step) which is slower than
the others and which therefore controls the rate of reaction.
If the rate-determining step is the first or the only in a sequence, then the
rate law can be written down directly from the stoichiometry of the first
step; e.g. the overall reaction
N2O5(g) + NO(g) -----> 3 NO2(g)
There are only two reactants and it is a one-step reaction. Therefore the
rate law can be determined to be
rate = k [N2O5][NO]
In an reaction that is not a one-step reaction then you must first determine
what the slowest step in the mechanism will be. Then once you've decided
on the slowest step, the rate law can be written directly from the slowest
step.