Chapter 14


Kinetics is the study of how fast chemical
reactions occur.
There are 4 important factors which
affect rates of reactions:




reactant concentration,
temperature,
action of catalysts, and
surface area.


Speed of a reaction is measured by the change
in concentration with time.
For a reaction A  B
change in number of moles of B
Average rate 
change in time
moles of B

t
At t = 0 (time zero) there
is 1.00 mol A (100 red
spheres) and no B
present.
 At t = 20 min, there is
0.54 mol A and 0.46 mol
B.
 At t = 40 min, there is
0.30 mol A and 0.70 mol
B.
moles of B 
Average rate 
t
moles of B at t  10  moles of B at t  0 

10 min  0 min
0.26 mol  0 mol

 0.026 mol/min
10 min  0 min


For the reaction A  B there are two
ways of measuring rate:


the speed at which the products appear (i.e.
change in moles of B per unit time), or
the speed at which the reactants disappear
(i.e. the change in moles of A per unit time).
moles of A 
Average rate with respect to A  
t
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
We can calculate the average rate in terms of
the disappearance of C4H9Cl.
 The units for average rate are mol/L·s or M/s.
 The average rate decreases with time.
 We plot [C4H9Cl] versus time.
 The rate at any instant in time (instantaneous
rate) is the slope of the tangent to the curve.
 Instantaneous rate is different from average
rate.
 We usually call the instantaneous rate the rate.




For the reaction:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
we know
C4H9Cl C4H9OH
Rate  

t
t
In general for:
aA + bB  cC + dD
1 A
1 B 1 C 1 D
Rate  



a t
b t
c t
d t

In general rates increase as concentrations
increase.
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)


For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
we note



as [NH4+] doubles with [NO2-] constant the rate
doubles,
as [NO2-] doubles with [NH4+] constant, the rate
doubles,
We conclude rate  [NH4+][NO2-].

Rate law:

The constant k is the rate constant.
Rate  k[ NH 4 ][ NO2 ]

For a general reaction with rate law
m
Rate  k[reactant 1] [reactant 2]n

we say the reaction is mth order in reactant 1
and nth order in reactant 2.

The overall order of reaction is m + n + ….


A reaction can be zeroth order if m, n, … are zero.
Note the values of the exponents (orders) have
to be determined experimentally. They are not
simply related to stoichiometry.




A reaction is zero order in a reactant if the
change in concentration of that reactant
produces no effect.
A reaction is first order if doubling the
concentration causes the rate to double.
A reacting is nth order if doubling the
concentration causes an 2n increase in rate.
Note that the rate constant does not depend on
concentration.

For a first order reaction, the rate doubles as
the concentration of a reactant doubles.
[A]
Rate  
 k[A]
t
ln A t  ln A 0  kt
 A t 
   kt
ln 
 A 0 


A plot of ln[A]t versus t is a straight line with
slope -k and intercept ln[A]0.
In the above we use the natural logarithm, ln,
which is log to the base e.
ln At  kt  ln A0

For a second order reaction with just one
reactant
1
1
 kt 
At
A0


A plot of 1/[A]t versus t is a straight line with
slope k and intercept 1/[A]0
For a second order reaction, a plot of ln[A]t vs. t
is not linear.



Half-life is the time taken for the concentration
of a reactant to drop to half its original value.
For a first order process, half life, t½ is the time
taken for [A]0 to reach ½[A]0.
Mathematically,
t1  
2
2   0.693
ln 1
k
k

For a second order reaction, half-life depends in
the initial concentration:
t1
2
1

k A0

Most reactions speed up as temperature
increases.

Examples
Food spoils when not refrigerated.
 When two light sticks are placed in water: one at
room temperature and one in ice, the one at room
temperature is brighter than the one in ice.
 The chemical reaction responsible for
chemiluminescence is dependent on temperature:
the higher the temperature, the faster the reaction
and the brighter the light

• As temperature
increases, the
rate increases.


Since the rate law has no temperature
term in it, the rate constant must depend
on temperature.
Consider the first order reaction CH3NC
 CH3CN.

As temperature increases from 190 C to 250 C
the rate constant increases from 2.52  10-5 s-1
to 3.16  10-3 s-1.

The collision model: in order for molecules to
react they must collide.
The greater the number of collisions the faster the
rate.
 The more molecules present, the greater the
probability of collision and the faster the rate.
 The higher the temperature, the more energy
available to the molecules and the faster the rate.


Complication: not all collisions lead to products.
In fact, only a small fraction of collisions lead to
product.

In order for reaction to occur the reactant molecules
must collide in the correct orientation and with
enough energy to form products.
Cl + NOCl  NO + Cl2
•
Arrhenius: molecules must posses a
minimum amount of energy to react.
–
–
•
In order to form products, bonds must be
broken in the reactants.
Bond breakage requires energy.
Activation energy, Ea, is the minimum
energy required to initiate a chemical
reaction.

Consider the rearrangement of methyl isonitrile:
N
H3C N C
H3C
H3C C N
C



In H3C-NC, the C-NC bond bends until the C-N
bond breaks and the NC portion is
perpendicular to the H3C portion. This
structure is called the activated complex or
transition state.
The energy required for the above twist and
break is the activation energy, Ea.
Once the C-N bond is broken, the NC portion
can continue to rotate forming a C-CN bond

From kinetic molecular theory, we know
that as temperature increases, the total
kinetic energy increases.

We can show the fraction of molecules, f,
with energy equal to or greater than Ea is
f e

E
 a
RT
where R is the gas constant (8.314
J/mol·K).

Arrhenius discovered most reaction-rate
data obeyed the Arrhenius equation:
k  Ae




 Ea
RT
k is the rate constant, Ea is the activation
energy, R is the gas constant (8.314 J/K-mol)
and T is the temperature in K.
A is called the frequency factor.
A is a measure of the probability of a favorable
collision.
Both A and Ea are specific to a given reaction.

If we have a lot of data, we can determine Ea
and A graphically by rearranging the Arrhenius
equation:
Ea
ln k  
 ln A
RT

From the above equation, a plot of ln k versus
1/T will have slope of –Ea/R and intercept of ln
A.

If we do not have a lot of data, then we
recognize
Ea
Ea
ln k1  
 ln A and ln k2  
 ln A
RT1
RT2
 Ea
  Ea

ln k1  ln k2   
 ln A    
 ln A 
 RT1
  RT2

k1 Ea  1 1 
ln 
  
k2 R  T2 T1 



The balanced chemical equation provides
information about the beginning and end of
reaction.
The reaction mechanism gives the path of the
reaction.
Mechanisms provide a very detailed picture of
which bonds are broken and formed during the
course of a reaction.


Elementary step: any process that occurs in a
single step.
Molecularity: the number of molecules
present in an elementary step.



Unimolecular: one molecule in the elementary
step,
Bimolecular: two molecules in the elementary
step, and
Termolecular: three molecules in the
elementary step.
 It is not common to see termolecular processes
(statistically improbable).


Some reaction proceed through more than one
step:
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
Notice that if we add the above steps, we get
the overall reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)

If a reaction proceeds via several elementary steps,
then the elementary steps must add to give the
balanced chemical equation.

Intermediate: a species which appears in an elementary
step which is not a reactant or product.

Rate Laws for Elementary Steps

The rate law of an elementary step is
determined by its molecularity:




Unimolecular processes are first order,
Bimolecular processes are second order, and
Termolecular processes are third order.
Rate Laws for Multistep Mechanisms


Rate-determining step: is the slowest of the
elementary steps.
Therefore, the rate-determining step governs
the overall rate law for the reaction.
2NO(g) + Br2(g)  2NOBr(g)
The experimentally determined rate law is
Rate = k[NO]2[Br2]
Consider the following mechanism
Step 1: NO(g) + Br2(g)
Step 2: NOBr2(g) + NO(g)
k1
k-1
k2
NOBr2(g)
(fast)
2NOBr(g) (slow)
The rate law is (based on Step 2):
Rate = k2[NOBr2][NO]


The rate law should not depend on the concentration
of an intermediate (intermediates are usually
unstable).
Assume NOBr2 is unstable, so we express the
concentration of NOBr2 in terms of NOBr and Br2
assuming there is an equilibrium in step 1 we have
k
[ NOBr2 ]  1 [ NO][Br2 ]
k1

By definition of equilibrium:
k1[ NO][Br2 ]  k1[ NOBr2 ]

Therefore, the overall rate law becomes
k
k
Rate  k2 1 [ NO][Br2 ][ NO]  k2 1 [ NO]2[Br2 ]
k1
k1

Note the final rate law is consistent with the
experimentally observed rate law


A catalyst changes the rate of a chemical
reaction.
There are two types of catalysts:


homogeneous, and
heterogeneous.

The catalyst and reaction is in one phase.

Generally, catalysts operate by lowering the
activation energy for a reaction.

Catalysts can operate by increasing the number of
effective collisions.


That is, from the Arrhenius equation: catalysts increase k
be increasing A or decreasing Ea.
A catalyst may add intermediates to the reaction.
 The
catalyst and reactants are in different
phases.

Process
First step is adsorption (the binding of reactant
molecules to the catalyst surface).
 Adsorbed species (atoms or ions) are very reactive.
 Molecules are adsorbed onto active sites on the
catalyst surface.
