Chemical Kinetics
Basic Terms and Concepts
• Introduction
• Types and Rate of Reaction
• Factor affecting reaction rates
• Rate Law and Rate Constant
• Order of reaction
• Molecularity of reaction
• Basic kinetic laws (zero,1st, 2nd, 3rd order)
1
What is Kinetics ?
Kinetics is the study of the speed at which chemical
and physical processes take place.
Chemical kinetics deals with trying to answer the
following questions:
1. At what rate does a chemical reaction undergo
change under a given set of conditions ?
2. What effect will a change in conditions have on
the rate at which a chemical change occurs?
3. What information is available as to HOW the
chemical reaction occurs ?
2
Types of Reaction
Two groups of chemical reactions - homogenous and
heterogeneous.
Homogeneous - reaction occurs in one phase only and
the system is uniform throughout.
E.g. H2 (g) + I2(g) → 2HI(g)
HCl(l)
CH3COOCH3(l) + H2O(l) → CH3COOH(l) + CH3OH(l)
Heterogeneous - reaction on the surface of a catalyst or
the walls of a container; the mixture is not uniform
throughout.
tungsten
2NH3(g) → N2(g) + 3H2(g)
alumina
C2H5OH(l) → C2H4 (g) + H2O(l)
3
Types of Reaction Rates
There are two types of reaction rates:
1. Instantaneous Rates
2. Average Rates
Instantaneous Rates - the change in product formation
at a moment in time.
A plot of product concentration vs. time shows a curve with
the tangent slope at any place on that curve gives the
instantaneous rate at that moment in time.
tan α = slope = side opposite/side adjacent = δy/δx
4
Types of Reaction Rates
Average rate - the difference in the concentration
of product or reactant divided by the the difference
in the time.
Av. Rate = ([product]final - [product]initial)/(tfinal - tinitial)
Av. Rate = - ([reactant]final - [reactant]initial)/(tfinal - tinitial)
5
DEFINITIONS OF RATE
 Rate
of Consumption of a reactant is -d[R]/dt
 Rate of Formation of a product is d[P]/dt
6
Rate of Reaction
The rate of reaction is defined as the no. of molecules
of a given species reacting per unit time.
It shows how the concentration of the reactants or
products changes with time.
Example:
CO + NO2 → CO2 + NO
200°C
The rate of this reaction may be expressed in terms of
change in the concentration of reactants or products.
Rate = -d[reactant]/dt = -d[CO]/dt = -d[NO2]/dt
The minus sign makes the rate a positive quantity.
7
Rate of Reaction
The rate can also be expressed in terms of change
in the concentration of products.
Rate = + d[product]/dt = + d[CO2]/dt = + d[NO]/dt
The plus sign is associated with product because
its concentration increases.
8
9
RATE OF REACTION

Consider a reaction of the form
A + 2B → 3C + D

B is consumed twice as fast as A; C is formed thrice as
fast as D

Only one numerical value is used to represent the
whole reaction.
Rate of reaction, r = d[D]/dt = 1/3 d[C]/dt
= - d[A]/dt = -½ d[B]/dt
10
Rate of Reaction
Example:
2HI(g) → H2(g) + I2(g)
The rate of consumption of HI is double the rate of
formation of H2 or I2.
International convention:
∆[reactant or product]
Rate of reaction, r = —————————————
Appropriate stoichiometric factor
Time derivative of [substance]
= ——————————————
Appropriate stoichiometric factor
= - ½ d[HI]/dt = d[H2]/dt = d[I2]/dt
11
Reaction rate has the units of concentration divided by time.
Reaction Rates
Example: N2(g) + 3H2(g) → 2 NH3(g)
The rate of this reaction expresses how fast the
reactants N2 and H2 disappear and how fast the
product NH3 forms.
Rate = -d[N2]/dt = -1/3d[H2]/dt = + 1/2d[NH3]/dt
The negative signs indicate [N2] and [H2] decrease.
The positive sign indicates [NH3] increase with time.
12
Reaction Rates
Example:

The rate of consumption of CH3 radicals in the
reaction 2 CH3(g) → CH3CH3(g) was reported as 1.2
mol L-1 s-1 under a certain set of conditions. What
is the rate of formation of CH3CH3(g)?

There is 1 mol CH3CH3 formed for every 2 mol CH3
consumed.

Therefore the rate of formation of CH3CH3 is half
the rate of consumption of CH3, or 0.6 mol L-1 s-1
13
Factor Affecting Reaction Rates
1. Nature of the reactants
2. Surface area of the reactants
3. Concentration of reactants
4. Temperature
5. Nature of catalyst, if present.
14
Nature of Reactants
Reactions between polar ionic molecules are rapid
Example: Ag+ + Cl- → AgCl (s)
Electron transfer reactions take longer time than ionic
reactions.
Example: Fe2+ + Ce4+ → Ce3+ + Fe3+
(Ce4+ + e- → Ce3+)
(Fe2+ → Fe3+ + e-)
Complex reactions are slow in nature
Example: Mg2+ + C2O42- → MgC2O4(s)
15
Surface area of the Reactants
In heterogeneous reaction, the rate of reaction
depends on the area of contact between the phases.
The rate of reaction is proportional to the surface area.
Particle size of reactant ↓
=> surface area ↑
=> rate of reaction ↑
16
Concentration of Reactants
Raising the concentration of a reactant increases
the reaction rate.
For a reaction A → Product, rate is doubled when
[A] is doubled.
[Reactant] ↑ => No. of particles/volume ↑
=> Effective collision rate of molecule ↑
=> Higher rate of reaction.
17
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
0.3 M HCl
6 M HCl
CONCENTRATION AND RATE
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
If we compare Experiments 1 and 2, we see
that when [NH4+] doubles, the initial rate
doubles.
© 2009, Prentice-Hall, Inc.
CONCENTRATION AND RATE
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Likewise, when we compare Experiments 5
and 6, we see that when [NO2−] doubles, the
initial rate doubles.
© 2009, Prentice-Hall, Inc.
Rate Laws

A rate law is an equation expressing the rate of a reaction
in terms of the molar concentrations of the species involved
in the reaction.

The rate is often found to be proportional to the molar
concentrations of the reactants raised to a simple power.

These powers are the orders of the reaction.

For the rate expression, Rate = k[A]m[B]n;
then the reaction is m order in A, n order in B, and m+n
order overall.
21
Rate Laws
Basic Rate Expression
Consider the following reaction:
aA + bB → Products
The rate expression is:
Rate = k [A]m[B]n
where k = rate constant
m = rate order in respect to A
n = rate order in respect to B
The TOTAL Order of the reaction is m + n.
22
Rate Laws
At constant temperature, the rate of a reaction is a
function of the concentration of reactants or products.
The rate law for a reaction cannot be determined from
the balanced chemical equation for the reaction. It can be
determined only by experiment.
The rate law provides concise expressions for the course
of reaction, helps in calculating reaction times, yields,
and optimum conditions.
23
Rate Laws
Example:
 The
reaction between the amino acid
tyrosine (Tyr) and iodine obeys the rate law:
r = k[Tyr][I2] Classify it by order.
 From
the powers in the rate expression, (1
and 1), the reaction is first order in Tyr,
first order in I2, and second order overall.
24
Rate Constant
The rate constant of a chemical reaction is a measure
of the rate of reaction when all the reactants are at unit
concentrations.
Rate = k x (function of concentration of reactant)
If concentrations are equal to unity,
∴ Rate of reaction = k where k is rate constant or
rate coefficient.
25
Rate Constant
Rate constant, k depends on the nature of reaction
and temperature. => k increases with temperature.
The rate constant is a characteristic of a chemical
reaction - a numerical measure of how fast a reaction
can occur.
If moles per liter is used, units for the rate constant
mole per liter
————————
(s)(moles per liter)n
where n is the total order of reaction.
First order, unit of k is s-1;
26
2nd order, unit of k is liter mol-1s-1.
EXERCISE: The reaction,
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
EXERCISE : The reaction,
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
Rate(Ms-1)
= k [H2] [NO]
b) What is the overall order of the reaction?
3
EXERCISE : The reaction,
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
Rate(Ms-1)
= k [H2] [NO]
3
b) What is the overall order of the reaction?
Overall order =
1+3 =4
c) What are the units of the rate constant?
“4th order”
EXERCISE : The reaction,
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
is experimentally found to be first order in H2 and third order in NO
a) Write the rate law.
Rate(Ms-1)
= k [H2] [NO]
3
b) What is the overall order of the reaction?
Overall order =
1+3 =4
c) What are the units of the rate constant?
“4th order”
Balanced Equations & Rate Expression
H2(g) + I2(g) → 2HI(g)
Rate = k[H2][I2]
2 N2O5(g) → 4 NO2(g) + 2H2O(g) Rate = k[N2O5]1
2 NOCl(g) → 2 NO (g) + Cl2 (g)
Rate = k[NOCl]2
CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g)
Rate = k[CHCl3][Cl]1/2
31
ELEMENTARY REACTIONS
 An
elementary reaction is one whose mechanism of
occurrence is just what is written in the reaction
equation.
 Elementary
reactions involve only one or two
molecules.
Example: H + Br2 → HBr + Br
 This
equation signifies that an H atom attacks a Br2
molecule and produces an HBr molecule and a Br atom.
32
Molecularity vs. Order
Terms that seem to conflict
1. MOLECULARITY: represents the stoichiometric
values used to balance the chemical equation.
2. KINETIC ORDER: represents values determined
experimentally and cannot, in general deduced from
the coefficients of the balanced equation.
33
MOLECULARITY VS. ORDER
 The
molecularity of a reaction refers to a particular
elementary reaction which is postulated as part of a
proposed mechanism.
 The
order of a reaction is an empirical quantity
which follows from an experimentally-determined
rate law.
 For
an elementary reaction, the rate law can be
written down from its chemical equation.
34
MOLECULARITY
 The
molecularity of an elementary reaction is the
number of molecules coming together to react.
 Elementary


reactions may be
unimolecular, in which a single molecule shakes itself apart or
into a new arrangement of its atoms
bimolecular, in which two molecules collide and undergo
some changes.
35
WRITING RATE LAWS FOR ELEMENTARY
REACTIONS

The reaction H + Br2 → HBr + Br is bimolecular.
is r = k [H][Br2]
Its rate law

Bimolecular reactions are second order. (But it is NOT the case that
second-order reactions are necessarily bimolecular, unless they are
elementary reactions.)

Unimolecular reactions are first order.

For a reaction such as
d[A]/dt = - ka[A]
d[I]/dt = ka[A] - kb[I]
d[P]/dt = kb[I]
A
ka
I
kb
.
P
36
Order of Reaction
Consider a general reaction
aA + bB + cC → pP + qQ + ….
Rate = k [A]x[B]y[C]z
The order of reaction is defined as the sum of the
powers to which the concentration terms are raised
in the rate-law equation.
The total order of reaction = x + y + z
37
Order of Reaction
Example: The decomposition of hydrogen peroxide
H2O2 → H2O + 1/2 O2
Rate Law:
-d[H2O2]/dt = k[H2O2]
First order reaction since the concentration
term is raised to the power of unity.
Example:
Rate Law:
2 CH3CHO → 2 CH4 + 2 CO
-d[CH3CHO]/dt = k[CH3CHO]2
38
Second order reaction
Order of Reaction
Example: The reaction 2 NO + O2 → 2 NO2
Rate Law: -d[NO]/dt = k[NO]2[O2]
Second order with respect to NO and First order with
respect to O2.
The overall reaction is thus of the Third order.
39
Order of Reaction
It is very seldom that reactions of order > 3 exist.
But there are reactions in which order of reaction may
be zero or fractional.
450°C
Example: CH3CHO
→
CH4 + CO
Rate expression: -d[CH3CHO]/dt = k[CH3CHO]3/2
The overall reaction is of the order of 3/2.
Zero order reaction - reaction which is independent
40
of concentration of reactions.
Effect of Concentration on Rate
Consider the reaction between nitric oxide and oxygen
2 NO(g) + O2(g) → 2 NO2(g)
Rate = k [NO]m [O2]n
Experimentally m =2, n =1;
The reaction is 2nd order in [NO]; 1st order in [O2]
or 3rd order overall.
But what does this really mean ?
41
Effect of Concentration on Rate
Rate = k [NO]m[O2]n
If [NO] is held constant in 2 experiments while
[O2] is doubled from c to 2c (Isolation Method).
The rate law predicts that the rate will double.
Rate (exp 2)
k[NO]2 [2c]
————— = —————— = 2
Rate (exp 1)
k[NO]2 [c]
If [O2] is doubled, the rate will double (first order).
42
Effect of Concentration on Rate
Rate = k [NO]m[O2]n
If [O2] is held constant in 2 experiments while
[NO] is doubled from c to 2c (Isolation Method)
=> the rate law predicts that the rate will quadruple.
Rate (exp 2)
k [2c]2 [O2]
————— = —————— = 22 = 4
Rate (exp 1)
k [c]2 [O2]
The fact that the reaction is second order in NO
=> the rate is proportional to the square of [NO]
43
Effect of Concentration on Rate
(General Rules – Isolation Method)
If the concentration of one reagent doubles while the
other reagent concentration is unchanged, and the rate is:
Unchanged - the order is ZERO with respect to the
changing reactant.
Doubled - the order is FIRST with respect to the
changing reactant.
Quadrupled - the order is SECOND with respect to the
changing reactant.
44
Determining the Rate Law
Exercise
Consider the reaction: CO + NO2 → CO2 + NO
[CO] M 0.10 0.20 0.30 0.10 0.20 0.30
[NO2]M 0.10 0.10 0.10 0.20 0.20 0.20
Rate M/s 0.005 0.010 0.015 0.010 0.020 0.030
What is the rate of the reaction with respect to both
CO and NO2 ?
45
Determining the Rate Law
Methods of Initial Rates
The method of initial rates measures the rate at the
beginning of a reaction for several different initial
concentrations of reactant.
46
Determining the Rate Law
Method of Initial Rates
Example
The initial rate of a reaction depended on the concentration of
a substance J as follows.
5.0
8.2
17
30
[J]0/(10-3 M)
-7
-1
r0 /(10 M s )
3.6
9.6
41
130
Find the order of the reaction with respect to J and the rate
constant.
r0 = k[J]0n, so log
r0 = n log [J]0 + log k
log [J]0
0.70 0.91 1.23 1.48
log r0
0.56 0.98 1.61 2.11
log r0 goes up twice as fast as log [J]0, so n = 2 and
log k = 0.56-2(0.70) = -0.84; k = 0.15 M-1s-1
47
Integrated Rate Laws
The concentration of a reactant, A, as a function of time,
can be equated to the rate law, r = k[A]n or
- dA/dt = k[A]n
Zero Order Kinetics
A reaction of zero order is represented as A → B
where A and B are the reactant and product respectively.
-d[A]/dt = k[A]0 = k or -d[A] = - kdt
Integrating between [A]0 and [A] and between 0 and t,
[A]t - [A]0 = - kt ----- (1), and rearranging
[A]0 - [A]t = kt ------- (2)
48
Zero Order Kinetics
A Zero Order reaction is one whose rate does
not depend on the concentration of the reactants.
i.e. Rate = k.
Rearranging equation (2) into the
following form:
[A] = -kt + [A]0
[A]
---- (3)
A plot of [A] vs. Time gives
a straight line for zero order
reactions.
49
Time
Zero Order Kinetics
Ways to Determine if a reaction is Zero Order
(1) Solve equation (2) for the rate constant k
k = ([A]0 - [A])/t
Insert [A] at various times, t. If the value of the rate
constant , k is constant, then the reaction is ZERO
ORDER.
The units for a zero order rate constant are
moles/liter-time (e.g. moles/liter-sec)
50
Zero Order Kinetics
Ways to determine if a reaction is Zero Order
(2) Half-Life Method:
The time for a reaction to go half-way is defined
by: t = t1/2; [A] = [A]0/2
Equation (2) becomes:
[A]0 - [A]0/2 = kt1/2
Solving for t1/2 : t1/2 = [A]0/2k
The Half-Life is directly proportional to the initial
concentration of reactant.
51
Characteristics of Zero Order Kinetics
1. The dimension of specific rate constant is
concentration/time, i.e. mol l-1 s-1
2. The product concentration is directly proportional
to the time.
=> plot of product (x) vs time (t) for such a reaction
would give a straight line passing through the origin,
and the gradient is the rate constant
3. The half-life of the zero order reaction is proportional
to the initial concentration of reactant.
When x = a/2, t = t1/2 => a/2 = k0t1/2 => t1/2 ∝ a
52
where a is the initial concentration of the reactant.
Examples of Zero Order Reaction
Some heterogeneous reactions of zero order occurring
on surfaces are:
hot Pt wire
2 N2 O
→
2 N2 + O2
tungsten
2 NH3
→
N2 + 3H2
gold
2 HI
→ H2 + I2
The reaction occurs only on the surface of the catalyst.
When the surface is saturated, the amount of absorbed
gas is constant
=> increase in pressure cannot change the surface conc.53
=> reaction rate is independent of [reactant].
Examples of Zero Order Reaction
In reactions between two or more substances,
the rate may have a zero order dependence upon
the concentration of one of them.
Example: CH3COCH3 + I2 → ICH2COCH3 + HI
Rate = -d[I2]/dt = k[CH3COCH3 ]
The reaction is first order in acetone, but zero order
in iodine.
The rate may be measured by monitoring the decrease
54
in [I2].
First Order Kinetics
Many reactions follow First Order Kinetics.
The general reaction is: A → Products
Rate = - d[A]/dt = k[A]1
Rearranging the equation:
-d[A]/[A] = kdt
----- (1)
continue...
55
First Order Kinetics
- d[A]/[A] = kdt
----- (1)
Taking equation (1) between the limits of [A] = [A]0
at t = 0 and [A] = [A] at t = t {integrate between limits}
ln [A] - ln [A]0 = - kt
Rearranging and combine the log terms:
ln([A]0/[A]) = kt -------- (1A)
56
First Order Kinetics
Ways to determine if a Reaction is First Order (1)
Rearranging equation 1A,
ln [A] = -kt + ln [A]0 ------- (3)
y
= mx + c
A plot of ln[A] vs time
gives a straight line for
First Order Reactions
ln [A]
57
time
First Order Kinetics
Ways to determine if a Reaction is First Order (2)
ln([A]0/[A]) = kt -------- (1A)
Solve this equation for the rate constant k;
k = (1/t)ln ([A]0 / [A]) ------- (2)
Insert [A] at various time t. If the value of rate
constant k is constant, then the reaction is FIRST order.
The units for a First Order Rate constant are 1/time
(e.g. sec-1).
58
First Order Kinetics
Ways to determine if a Reaction is First Order (3)
Half-Life Method: the time required for 1/2 of
reactant to react.
At t = t1/2, equation (1A) becomes:
ln([A]0/{ln[A]0/2}) = kt1/2
Solving for t1/2: t1/2 = ln 2/ k = 0.693/k
59
Half-life is INDEPENDENT of the initial concentration
!
First Order Kinetics
Rate = - d[A]/dt = k[A]1
An example of a first order reaction is the
decomposition of nitrogen pentoxide.
2 N2O5(g) → 4 NO2(g) + O2(g)
Rate = -d[N2O5]/dt = k[N2O5]1
Note: Kinetic Order and Molecularity are not
the same in this example.
60
First Order Kinetics
Examples of First Order Processes
1. All nuclear process follow First Order Kinetics
a. U238 → Th 232 + α
b. C14 → N14 + β
2. Decomposition Reactions:
H2C2O4 → CO2 + HCOOH
61
Rate = k[H2C2O4]1
Exercise:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50
mg remains after 4.26 min, what is the rate constant in s−1?
Exercise:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50
mg remains after 4.26 min, what is the rate constant in s−1?
Begin with the integrated rate law for a 1st order process:
[A]t = [A]o e
− kt
[A]t
= e−kt
[A]o
[N2O5 ]t
[N2O5 ]o
Wait… what is the volume of the container???
Do we need to convert to moles?
Exercise:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50
mg remains after 4.26 min, what is the rate constant in s−1?
[N2O5 ]t
[N2O5 ]o
Check it out!
[N2O5 ]t
[N2O5 ]o
You don’t need the
volume of the container!
1g
1 mol
2.50 mg × 3
×
10 mg 108.0 g
2.50 mg
x
L
=
1g
1 mol
2.56
mg
2.56 mg × 3
×
10 mg 108.0 g
xL
Exercise:
The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50
mg remains after 4.26 min, what is the rate constant in s−1?
[N2O5 ]t
=
[N2O5 ]o
2.50 m
e=
2.56 m
− kt
taking the natural log and substituting time in seconds:
 2.50 mg 
ln 
=−k × ( 256 s )

 2.56 mg 
k = 9.3 ×10−5 s−1
Pseudo First Order Reactions
Some reactions which obey the 1st order kinetics
but involve more than one species in the ratedetermining step.
They involve a solvent molecule or a catalyst as
one of the reacting species.
1. Hydrolysis of methyl acetate
H+
CH3COOCH3 + H2O → CH3COOH + CH3OH
2. Inversion of sucrose
H+
C12H22O11 + H2O → C6H12O6 + C6H12O6
66
Pseudo First Order Reactions
In these cases, water is present in large excess,
=> its concentration virtually remains constant
during the course of reaction.
=> the rate of reaction depends on [CH3COOCH3]
or sucrose.
-d[CH3COOCH3]/dt = k1[CH3COOCH3][H2O]
Since [H2O] = b is constant.
∴ -d[A]/dt = k1[A].b
Upon integration, a new rate constant can be defined
k = b.k1 = 1/t (ln[A]0/[A])
67
Second Order Kinetics
Many Reactions such as: 2 HI → H2 + I2
are second order kinetics.
The general format for such reactions is:
2A → Products
Rate = k [A]2
----- (1)
68
Second Order Kinetics
Rate = -d[A]/dt = k [A]2
Rearranging: -d[A]/[A]2 = k dt
Integration from [A] =[A]0 at t =0 to
[A] = [A] at t = t
1/[A] - 1/[A]0 = kt
------ (2)
69
Second Order Kinetics
Ways to Determine if a Reaction is Second Order (1)
Rearranging: 1/[A] - 1/[A]0 = kt
1/[A] = kt + 1/[A]0
1/[A]
A plot of 1/[A] vs. Time
gives a straight line with a
slope = k.
Time
70
Second Order Kinetics
Ways to Determine if a Reaction is Second Order (2)
1/[A] - 1/[A]0 = kt ------ (2)
Solving for k from equation (2)
k = (1/t){1/[A] - 1/[A]0} ------(2A)
Insert [A] at various time t. If the value of rate constant,
k is constant, then the reaction is SECOND ORDER.
The units for a Second Order Reaction are
1/(Moles/liter)(time) or Liters/mole-time
71
Second Order Kinetics
Ways to Determine if a Reaction is Second Order (3)
Half-Life Method: the time required for half of
reactant to react.
At t = t1/2, [A] = [A]0/2
Substituting into: 1/[A] - 1/[A]0 = kt ------ (2)
1/([A]0/2) - 1/[A]0 = kt1/2
Solving for t1/2:
t1/2 = 1/k[A]0
The half-life for a second order reaction is inversely
proportional to the initial concentration.
72
Second Order Kinetics
Characteristics of Second Order Kinetics
1. The unit for a second order rate constant are
(molar concentration x time)-1 or 1 mol-1s-1.
2. A plot of 1/[A] vs. time gives a straight line
with a slope = k.
3. The half-life for a second order reaction is
inversely proportional to the first power of initial
concentration.
73
Examples of Second Order Reactions
1. Saponification of an ester is a 2nd order reaction.
CH3COOC2H5 + OH- → CH3COO- + C2H5OH
2. Reaction between ethylene bromide and potassium
iodide in 99% methanol.
C2H4Br2 + 3KI → C2H4 + 2KBr + KI3
3. Reaction between potassium persulphate and
potassium iodide in aqueous medium
K2S2O8 + 2KI → 2K2SO4 + I2
74
EXERCISE
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) + 12 O2 (g)
and yields data comparable to
this:
Time (s)
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
© 2009, Prentice-Hall, Inc.
76
• Plotting ln [NO2] vs. t yields
the graph at the right.
• The plot is not a straight line,
so the process is not firstorder in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
0.01000
−4.610
50.0
0.00787
−4.845
100.0
0.00649
−5.038
200.0
0.00481
−5.337
300.0
0.00380
−5.573
© 2009, Prentice-Hall, Inc.
1
[NO2vs.
]
• Graphing ln
t, however, gives this
plot.
Time (s)
[NO2], M
1/[NO2]
0.0
0.01000
100
50.0
0.00787
127
100.0
0.00649
154
200.0
0.00481
208
300.0
0.00380
263
• Because this is a
straight line, the
process is secondorder in [A].
© 2009, Prentice-Hall, Inc.
Higher Order Kinetics
Third and higher order reactions are quite rare.
This is due to the fact that the probability of a collision
between three molecules having sufficient energy is
very small compared with bimolecular collision.
Fractional Order of Kinetics
A reaction is said to be of fractional orders if the rate
of reaction is proportional to the fractional powers of
concentration of the reactants.
79
Summary
Summary of kinetic relationships for orders 0,1,and 2
Order Rate Law Integrated Form
Units of k
M.s-1
t1/2
0
R=k
[A]=-kt + [A]0
[A]0/2k
1
R = k[A]
ln[A]=-kt + ln[A]0 s-1
2
R = k[A]2
1/[A] = kt + 1/[A]0 M-1.s-1 1/k[A]0
0.693/k
80