Chapter 4 Chemical Kinetics Part 1 free study material by TEACHING CARE online tuition and coaching classes
The branch of physical chemistry which deals with the rate at which the chemical reactions occur, the mechanism by which the chemical reactions take place and the influence of various factors such as concentration, temperature, pressure, catalyst etc., on the reaction rates is called the chemical kinetics.
On the basis of reaction rates, the chemical reactions have been classified into the following three groups :
 Very fast or instantaneous reactions : These reactions occur at a very fast rate generally these reactions involve ionic species and known as ionic
These reactions take about 10^{–14} or 10^{–16} seconds for completion. It’s rate can be measured by employing special methods. So, it is almost impossible to determine the rates of these reactions.
Examples : (i)
(ii)
AgNO3 + NaCl ® AgCl+ NaNO3
(PPt.)
HCl+ NaOH ® NaCl+ H _{2}O
(Precipitation reaction)
(Neutralization reaction)
(acid)
(base)
(Salt)
 Moderate reaction : These type of reactions proceed with a measurable rates at normal temperature. In this a large number of bonds have to be broken in reactants molecules and a large number of new bonds have to be formed in product Mostly these reactions are molecular in nature.
Examples : (i) Decomposition of
 Decomposition of
H 2 O2 :
N 2 O5 :
2H 2 O2 ® 2H 2 O + O2
2N 2 O5 ® 2N 2 O4 + O2
 Hydrolysis of ester :
 Reaction of NO with chlorine :
CH3 COOC2 H5 + NaOH ® CH3 COONa + C2 H5 OH
NO + Cl_{2} ® NOCl_{2}
 Inversion of cane sugar in aqueous solution :
C12 H22O11 + H2O ® C6 H12O6 + C6 H12O6
2FeCl3 (aq.) + SnCl2 ® 2FeCl2 (aq.) + SnCl4 (aq.)
NO_{2} + CO ® NO + CO_{2}
Glucose
Fructose
 Decolourisation of acidified potassium permanganate with sodium
 Very slow reactions : These reactions are extremely slow and take months together to show any measurable The rate of such type of reactions are very slow. So, it is also very difficult to determine the rate of these reactions.
Examples : (i) Rusting of iron :
Fe 2 O3 + xH 2 O ®
Fe 2 O3 . xH 2 O
Hydrated ferric oxide (Rust)
 Reaction between H 2
and O_{2} to form
H 2 O
at ordinary temperature in absence of catalyst.
 Reaction of atmospheric
H _{2} S on basic lead acetate paint.
White basic lead acetate paint ¾¾atmo¾sphe¾ric ® Blackening of paint occurs very slowly
H 2 S
(due to formation of Pbs)
Note : ® The chemical reactions can be slowed down or speed up by changing conditions under which
they occur. e.g.
CO + 2H 2 ¾¾^{At} ^{r}¾^{oo}¾^{m} ® CH3 OH (very slow reaction)
temp.
The reaction can be speeded up by maintaining temperature around 400°C, pressure about 300 atm and using a catalyst containing ZnO and Cr2O3 .
“The rate (speed or velocity) of reaction is the rate of change in concentration of reactants or products in unit time.”
When, After,
t = 0
t = t
A ¾¾® Product a 0
(a–x) x
Where a is the initial concentration and (a–x) is concentration of reactant after time t and x will be the concentration of product after time t.
If dx is the change in concentration in time interval dt then,
The reaction rate for reactants = – dx ; The reaction rate for products = + dx
dt dt
 The negative sign indicates that the concentration of reactant decreases with
 The positive sign indicates that the concentration of products increases with
 The concentration change may be positive or negative but the rate of reaction is always positive.
 The rate of chemical reaction decreases as the reaction
 The concept of mechanical speed or velocity can not be used in measuring rate of Rate of reaction depends on molar concentration.
 Types of rate of reactions : There are two types of rate of
 Average rate of reaction : The average rate is defined as the change in the concentration (active mass) of reactants or products over a long time
Consider the general chemical reaction, aA + bB + …….. ¾¾® c C + dD + ……….
Average rate = Amount of reactant consumed (or product formed)/time interval.
Average rate = – D[A] = – D[B] ……….. = + D[C] = + D[D] = + ……..
aDt
bDt
cDt
dDt
as Dt
The average rate over the time interval Dt
approaches zero.
approaches the instantaneous rate
 Instantaneous rate of reaction : The instantaneous rate of reaction
gives the tendency of the reaction at a particular instant. The term Dt becomes
smaller and eventually approaches zero, then the rate of reaction at a particular moment called the instantaneous rate (Rt ) is given by,
Instantaneous rate = (Average rate) Dt ®0
R = æ – D[A] ö
= æ D[B] ö
or R
= – d[A] = d[B]

t ç Dt
÷

øDt ®0
ç Dt
÷ t
øDt ®0
dt dt
Where,
d[A], d[B]
and dt being infinitesimally small changes in the concentration of A and B , that of time
respectively. Instantaneous rate of reaction at any instant of time is obtained by finding the slope of the tangent to the curve (which is obtained by plotting concentration of any suitable reactant or product versus time) at the point
corresponding to that instant of time. Rate of reaction =
tanq = dx
dt
 Unit of rate of reaction : Unit of rate of reaction =
Unit of concentration = mole litre –1 time –1
Unit of time
 If reactants and products are in gaseous state then the pressure may be taken in place of concentration thus rate will have unit of atm sec^{–1} or atm min^{–1}
 The unit of time can be second, minute, hours, days and years so the unit of rate of reaction may be expressed as
follows: mol/litre sec ( mol l 1 s 1 )
or mol/litre min (
mol l ^{1} min ^{1} )
or mol/litre hour (mol l ^{1}h^{1} ) or mol/litre day
(mol l ^{1}d ^{1} ) or mol/litre year (mol l ^{1} y ^{1} )
Example : 1 For the reaction
N 2 + 3H 2
⇌ 2NH _{3} , if
D[NH 3 ] = 2 ´ 10 ^{–}^{4} mol l ^{–}^{1} s ^{–}^{1} , the value of
Dt
D[H 2 ]
Dt
would be
[MP PMT 2000]
(a)
1 ´ 10 ^{–}^{4} mol l ^{–}^{1}s ^{–}^{1}
(b)
3 ´ 10 ^{–}^{4} mol l ^{–}^{1}s ^{–}^{1}
(c)
4 ´ 10 ^{–}^{4} mol l ^{–}^{1}s ^{–}^{1}
(d)
6 ´ 10 ^{–}^{4} mol l ^{–}^{1}s ^{–}^{1}
Solution: (b)
N 2 + 3H 2 ⇌ 2NH 3
D[N _{2} ] = – 1 D[H _{2} ] = 1 D[NH _{3} ] ; \ D[H _{2} ] = 3 ´ D[NH _{3} ] = 3 ´ 2 ´ 10 4
= 3 ´ 10 ^{–}^{4} mol l ^{–}^{1}s ^{–}^{1} .
Dt 3 Dt
2 Dt
Dt 2 Dt 2
Example : 2 A gaseous hypothetical chemical equation 2A ⇌ 4B + C
is carried out in a closed vessel. The concentration of
B is found to increase by 5 ´ 10 ^{–}^{3} mol l ^{–}^{1}
in 10 second. The rate of appearance of B is [AFMC 2001]
(a)
(c)
5 ´ 10 ^{–}^{4} mol l ^{–}^{1} sec ^{–}^{1}
6 ´ 10 ^{–}^{5} mol l ^{–}^{1} sec ^{–}^{1}
(b)
(d)
5 ´ 10 ^{–}^{5} mol l ^{–}^{1} sec ^{–}^{1}
4 ´ 10 ^{–}^{4} mol l ^{–}^{1} sec ^{–}^{1}
Solution: (a) Increase in concentration of
B = 5 ´ 10^{–}^{3} mol l ^{–}^{1} , Time = 10 sec.
Rate of appearance of B = Increase of concentration of B = 5 ´ 10^{–}^{3} mol l ^{–}^{1} = 5 ´ 104 mol l 1 sec 1 .
Time taken 10 sec
Example : 3 If 3 A ® 2B then the rate of reaction of + d(B) is equal to [CBSE 2002]
dt
(a)
+ 2 d(A)
(b)
– 1 d(A)
(c)
– 2 d(A)
(d)
– 3 d(A)
dt 3 dt
3 dt
2 dt
Solution: (c)
3 A ® 2B ; Rate = – 1 d [A] = 1 d [B] ;
\ + d [B] = – 2 d [A]
3 dt 2 dt
dt 3 dt
Example: 4 A gaseous reaction,
A_{2} (g) ® B(g) + 1 C(g)
2
; Shows increase in pressure from 100 mm to 120 mm in 5
minutes. The rate of disappearance of
A2 is
(a)
4 mm min ^{–}^{1}
(b)
8 mm min ^{–}^{1}
(c)
16 mm min ^{–}^{1}
(d)
2 mm min ^{–}^{1}
Solution: (b)
A_{2} (g) ® B(g) + 1 C(g)
2
100 0 0
100 – p p 1 p
2
100 – p + p + 1 p = 120 or
2
p = 40mm
\ – dp A2
dt
= 40 = 8 mm min ^{–}^{1}
5
Many physical and chemical methods are available for studying the reaction rate :
 Volume or Pressure measurement : The reaction rate can be followed by measuring the volume or pressure change provided one or more of the components are
 Titrimetry : The reaction course can be followed using acidbase or oxidationreduction titration if at least one of the components in the reaction is an acid or a base or an oxidising agent or a reducing
 Conductometry or Potentiometry : It is a suitable method based on conductivity or potentiometric measurements if one or more of the ions are present or produced in the
 Spectrophotometry : When a component of the reaction has a strong absorption band at a particular wavelength region, spectrophotometers could be used for measuring the reaction
 Polarimetry : The reaction rate can be studied from the measurements of optical rotation when at least one of the component of a reaction is optically
The rate of a chemical reaction depends on the rate of encounter between the molecules of the reactants which in turn depends on the following things.
 Effect of temperature on reaction rate : The rate of chemical reaction generally increases on increasing the
 Nature of reactants : (i) Reactions involving polar and ionic substances including the proton transfer reactions are usually very fast. On the other hand, the reaction in which bonds is rearranged, or electrons transferred are
 Oxidationreduction reactions, which involve transfer of electrons, are also slow as compared to the ionic
 Substitution reactions are relatively much
 pH of the medium : The rate of a reaction taking place in aqueous solution often depends upon the H ^{+} ion Some reactions become fast on increasing the H^{+} ion concentration while some become slow.
 Concentration of reactants : The rate of a chemical reaction is directly proportional to the concentration of the reactants means rate of reaction decreases with decrease in
 Surface area of reactant : Larger the surface area of reactant, the probability of collisions on the surface of the reactant particles by the surrounding molecules increases and thus rate of reaction
 Presence of catalyst : The function of a catalyst is to lower
down the activation energy. The greater the decrease in the activation energy caused by the catalyst, higher will be the reaction rate. In the presence of a catalyst, the reaction follows a path of lower activation energy. Under this condition, a large number of reacting molecules are able to cross over the energy barrier and thus the rate of reaction increases. Fig. shows how the activation energy is lowered in presence of a catalyst.
 Effect of sunlight : There are many chemical reactions whose rate are influenced by radiations particularly by ultraviolet and visible light. Such reactions are called photochemical reactions. For example, Photosynthesis, Photography, Blue printing, Photochemical synthesis of compounds

H + Cl ¾¾sunli¾ght (¾h¾n ) ® 2HCl : The radiant energy initiates the chemical reaction by supplying the necessary activation energy required for the reaction.
 Rate law : The actual relationship between the concentration of reacting species and the reaction rate is determined experimentally and is given by the expression called rate
For any hypothetical reaction, aA + bB ® cC + dD

Rate law expression may be,
Where a and b are constant numbers or the powers of the concentrations of the reactants A and B
respectively on which the rate of reaction depends.
 Rate of chemical reaction is directly proportional to the concentration of the
 The rate law represents the experimentally observed rate of reaction, which depends upon the slowest step of the
 Rate law cannot be deduced from the relationship for a given It can be found by experiment only.
 It may not depend upon the concentration of species which do not appear in the equation for the over all
 Law of mass action : (Guldberg and Wage 1864) This law relates rate of reaction with active mass or molar concentration of reactants. According to this law, “At a given temperature, the rate of a reaction at a particular instant is proportional to the product of the reactants at that instant raised to powers which are numerically equal to the numbers of their respective molecules in the stoichiometric equation describing the ”
Active mass = Molar concentration of the substance = Number of gram moles of the substance
= W / m = n
Volume in litres V V
Where W =
volume in litre.
mass of the substance, m is the molecular mass in grams, ‘n’ is the number of g moles and V is
Consider the following general reaction, m1 A1 + m2 A2 + m3 A3 ® Products Rate of reaction µ [A1 ]^{m}1 [A2 ]^{m}2 [A3 ]^{m}3
 Rate constant : Consider a simple reaction,
A ® B . If CA
is the molar concentration of active mass of A
at a particular instant, then,
dx µ C or
dt A
dx = kC dt A
; Where k is a proportionality constant, called velocity
constant or rate constant or specific reaction rate constant.
At a fixed temperature, if CA = 1 , then
“Rate of a reaction at unit concentration of reactants is called rate constant.”
 The value of rate constant depends on, Nature of reactant, Temperature and Catalyst
(It is independent of concentration of the reactants)
é litre ù n1
émol ù1n
 Unit of rate constant : Unit of rate constant = ê
ú ´ sec 1 or = ê
ú ´ sec 1
Where n =
order of reaction
ëmol û
ë litre û
Difference between Rate law and Law of mass action
It is an experimentally observed law. It is a theoretical law.
It depends on the concentration terms on which the rate of reaction actually depends
Example for the reaction, aA + bB ® Products
Rate = k [A]^{m}[B]^{n}
It is based upon the stoichiometry of the equation
Example for the reaction, aA + bB ® Products
Rate = k[A]^{a} [B]^{b}
Difference between Rate of reaction and Rate constant
Rate of reaction  Rate constant 
It is the speed with which reactants are converted into  It is proportionality constant. 
products.  
It is measured as the rate of decrease of the  It is equal to rate of reaction when the 
concentration of reactants or the rate of increase of  concentration of each of the reactants is unity. 
concentration of products with time.  
It depends upon the initial concentration of the  It is independent of the initial concentration of 
reactants.  the reactants. It has a constant value at fixed 
temperature. 
“The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction.”
Or
“The total number of molecules or atoms whose concentration determine the rate of reaction is known as order of reaction.”
Order of reaction = Sum of exponents of the concentration terms in rate law
xA + yB ® Products
By the rate law,
Rate = [A]^{x} [B[^{y} , then the overall order of reaction.
n = x + y , where x and y are the orders
with respect to individual reactants.
If reaction is in the form of reaction mechanism then the order is determined by the slowest step of mechanism.
2A + 3B ® A_{2} B_{3}
A + B ® AB(fast)
AB + B_{2} ® AB_{3} (slow)
AB_{3} + A ® A_{2} B_{3} (fast)
(Rate determining step)
(Here, the overall order of reaction is equal to two.)
An order of a reaction may be zero, negative, positive or in fraction and greater than three. Infinite and imaginary values are not possible.
 First order reaction : When the rate of reaction depends only on the one concentration term of
Examples : · A ® product
 H 2 O2
® H 2
O + 1 O
2 2
Exception :
 All radioactive reactions are first order
 Rate of growth of population if there is no change in the birth rate or death
 Rate of growth of bacterial culture until the nutrients are
H _{2}O, H ^{+} , OH ^{–} and excess quantities are not considered in the determining process of order.
Examples : · CH 3 COOC2 H 5 + H 2 O ® CH 3 COOH + C2 H 5 OH ; Order = 1; R = k [CH3 COOC2 H5 ]
 2A(excess) + B ® product ; Order = 1; R = k [B]
 2N 2 O5 ® 4 NO2 + O2 ; Order = 1;
 2Cl2 O7 ® 2Cl2 + 7O2 ; Order = 1;
R = k [N 2 O5 ]
R = k [Cl_{2}O_{7} ]
 (CH3 )3 – C – Cl + OH ^{–} ® (CH3 )3 C – OH + Cl ^{–} ; Order = 1;
 Velocity constant for first order reaction : Let us take the reaction
A ¾¾® Product
Initially t = 0 a 0
R = k [(CH_{3} )_{3} C – Cl]
After time t = t (a – x) x
Here,
‘ a‘
be the concentration of A at the starting and
(a – x)
is the concentration of A after time t i.e., x
part has been changed in to products. So, the rate of reaction after time t is equal to
dx µ (a – x) or
dt
integrated rate constant is,
dx = k(a – x) or
dt
dx
(a – x)
= k.dt
…..(i)
…..(ii)
…..(iii)
 Half life period of the first order reaction : when t = t1 / 2 ;
x = a , then eq. (ii) becomes
2
t = 2.303 log a ; t = 2.303 log a
1 / 2 k
10 æ a ö 1 / 2 k
10 a / 2
ç a – 2 ÷
è ø
t = 2.303 log
2 (Qlog 2 = 0.3010 ); \ t
= 2.303 ´ 0.3010
1 / 2 k 10 1 / 2 k
Half life period for first order reaction is independent from the concentration of reactant. Time for completion of n^{th} fraction,
 Unit of rate constant of first order reaction : k = (sec)1
 Second order reaction : Reaction whose rate is determined by change of two concentration terms is said to be a second order For example,
 CH 3 COOH + C2 H 5 OH ¾¾® CH 3 COOC2 H 5 + H 2 O



 S2O^{2}^{–} + 2I ^{–} ¾¾® 2SO^{2}^{–} + I
(i) Calculation of rate constant :
2 A ¾¾® product or
A + B ¾¾® product
When concentration of A and B are same.
A + B ¾¾® Product
Initially
t = 0 a a 0
After time t = t
(a – x)
(a – x) x
dx = k[A][B] = k [a – x][a – x]
dt
dx = k [a – x]2 ; Integrated equation is ;
dt
When concentration of A and B are taken different
A  +  B  ®  Product  
Initially t = 0
After time t = t 
a
(a – x) 
b
(b – x) 
0
x 
dx = k [a – x].[b – x] , Integrated equation is,
dt
æ a ö
a 1 ç 2 ÷ 1
 Half life period of the second order reaction : When t = t1 / 2 ;
x = 2 ; t1 / 2 =
ç ÷ =



k ç a ´ (a – a ÷ ka
è 2 ø
Half life of second order reaction depends upon the concentration of the reactants.
(iii) Unit of rate constant :
reaction)
k = mol^{1}^{D}^{n} lit ^{Δ} ^{n}^{1} sec^{1} ; Dn = 2 ,
k = mol 1 lit. sec1
(Where
Dn =
order of
 Third order reaction : A reaction is said to be of third order if its rate is determined by the variation of three concentration terms. When the concentration of all the three reactants is same or three molecules of the same reactant are involved, the rate expression is given as
3 A ¾¾® products or
A + B + C ¾¾®
products
(i) Calculation of rate constant :
dx = k(a – x)3 , Integrated equation is

dt
 Half life period of the third order reaction : Half life period =
inversely proportional to the square of initial concentration.
æ mol ö2
3 ;
2a 2 k
Thus, half life is
 Unit of rate constant : k =ç ÷ time1 or k = litre 2mol 2 time1
è litre ø
 Zero order reaction : Reaction whose rate is not affected by concentration or in which the concentration of reactant do not change with time are said to be of zero order For example,

 H + Cl ¾¾Sunl¾ig¾ht ® 2HCl
 Dissociation of HI on gold
 Reaction between acetone and
 The formation of gas at the surface of tungsten due to
 Calculation of Rate Constant : Let us take the reaction
A ¾¾® Product
Initially t = 0 a 0
dx = k[A]0 ,
dt
dx = k ;
dt
dx = k. dt
Integrated rate equation, substance.
k = x ; The rate of reaction is independent of the concentration of the reacting
t
(ii) Half life period of zero order reaction : When
t = t1 / 2 ;
x = a ;
2
t1 / 2
= a or
2k
t1 / 2 µ a ; The half life
period is directly proportional to the initial concentration of the reactants.
 Unit of Rate constant : k = mole ; Unit of rate of reaction = Unit of rate
lit. sec.
Note : ® In general, the units of rate constant for the reaction of n^{th} order are:
Rate = k[A]^{n}
mol L^{1} s
= k(mol L^{1} )^{n}
or k = (mol L^{1} )^{1}^{–}^{n} s ^{1}
Units of rate constants for gaseous reactions: In case of gaseous reactions, the concentrations are
expressed in terms of pressure in the units of atmosphere. Therefore, the rate has the units of atm per second. Thus, the unit of different rate constants would be:
(i) Zero order reaction : atm s 1
(iii) Second order reaction: atm^{1}s ^{1}
(ii) First order reaction : s ^{1}
 Third order reaction: atm^{2} s ^{1}
In general, for the gaseous reaction of n^{th} order, the units of rate constant are (atm)^{1–n}s^{–1}
Modified expressions for rate constants of some common reactions of first order
Examples of reactions having different orders
Examples  Rate Law  Order 
First order reaction  
2H2O2 ¾¾® 2H2O + O2  r = k [H_{2}O_{2}]  1 
C2H5Cl ¾¾® C2H4 + HCl  r = k [C_{2}H_{5}Cl]  1 
2N2O5 ¾¾® 4 NO2 + O2  r = k [N_{2}O_{5}]  1 
SO2Cl2 ¾¾® SO2 + Cl2  r = k [SO_{2}Cl_{2}]  1 
CH3COOC2H5 + H2O ¾¾® CH3COOH + C2H5OH  r = k [ester][H_{2}O]^{0}  1 
C12H22O11 + H2O ¾¾® C6 H12O6 + C6 H12O6  r = k [sugar][H_{2}O]^{0}  1 
All radioactive decay  r = k [radioactive species]  1 
Second order reactions  
NO + O_{3} ¾¾® NO_{2} + O_{2}  r = k [NO] [O_{3}]  2 
2NO_{2} ¾¾® 2NO + O_{2}  r = k [NO_{2}]^{2}  2 
H_{2} + I_{2} ¾¾® 2HI  r = k[H_{2}][I_{2}]  2 
CH3COOC2H5 + OH^{–} ¾¾® CH3COO^{–} + C2H5OH r = k [CH3CO2C2 H5 ][OH – ]  2  
C2H4 + H2 ¾¾® C2H6  r = k [C2H4 ][H2 ]  2 
2N2O ¾¾® 2N2 + O2  r = k [N _{2}O]^{2}  2 
2CH_{3}CHO ¾¾® 2CH_{4} + 2CO  r = k [CH_{3}CHO]^{2}  2 
Third order reactions
2NO + O_{2} ¾¾® 2NO_{2} 
r = k [NO]^{2}[O_{2}] 
3 
2NO + Br_{2} ¾¾® 2NOBr  r = k [NO]^{2}[Br_{2}]  3 
2NO + Cl_{2} ¾¾® 2NOCl  r = k [NO]^{2}[Cl_{2}]  3 
Fe ^{2}^{+} + 2I ^{–} ¾¾® FeI 2  r = k [Fe^{2}^{+}][I ^{–}]^{2}  3 
Zero order reactions
H 2 + Cl 2 ¾¾® 2HCl (over water) 
r = k [H2 ]^{0}[Cl2 ]^{0} 
0 
2NH3 ¾¾^{Pt} ® N2 + 3H2  r = k [NH_{3}]^{0}  0 
Fractional order reactions
Para H _{2} ¾¾® ortho H _{2} 
r = k [para H2]^{1.5} 
1.5 
CO + Cl _{2} ¾¾® COCl _{2}  r = k [CO]^{2}[Cl2]^{1} ^{/} ^{2}  2.5 
 Substitution method in integrated rate equation (Hit and Trial method)
 The method can be used with various sets of a, x and t with integrated rate
 The value of k is determined and checked for all sets of a, x and t .
 If the value of k is constant, the used equation gives the order of
 If all the reactants are at the same molar concentration, the kinetic equations are :
k = 2.303 log
t
a
10 (a – x)
(For first order reactions)
k = 1 é1 – 1 ù
(For second order reactions)
t êë a a – x úû
k = 1 é 1 – 1 ù
(For third order reactions)


2t ê(a – x)2 a 2 ú
 Half life method : This method is employed only when the rate law involved only one concentration

t µ a1n ; t = ka1n ;
log t1 / 2 = log k + (1 – n) log a , a plotted graph of
log t1 / 2 vs log a gives a straight line
with slope (1 – n), determining the slope we can find the order n . If half life at different concentration is given then.


1 1 (t )
æ a ön1

(t ) µ
; (t ) µ
; 1 / 2 1 = ç 2 ÷
1 / 2 1
n1 1
1 / 2 2
n1
2
(t1 / 2 )2
ç ÷
è 1 ø
log10 (t1 / 2 )1 – log 10 (t1 / 2 )2 = (n – 1) [log10 a2 – log10 a1 ]

; This relation can be used to determine order of reaction ‘n’
Plots of halflives Vs concentrations (t_{1/2} µ a^{1–n})
 Graphical method : A graphical method based on the respective rate laws, can also be
 If the plot of log(a – x) Vs t is a straight line, the reaction follows first order.
 If the plot of
1
(a – x)
Vs t is a straight line, the reaction follows second order.
 If the plot of 1
(a – x)2
Vs t is a straight line, the reaction follows third order.
 In general, for a reaction of nth order, a graph of
1
(a – x)n1
Vs t must be a straight line.
Graphical determination of order of the reaction
Order  Equation  Straight line plot Slope Intercept on  
Y axis  Xaxis  Yaxis  
Zero  x = k0 t
log (a – x) = –k1t + log ^{10} 2.303 ^{10} (a – x)^{–}^{1} = k2 t + a ^{–}^{1} (a – x)^{1}^{–}^{n} = (n – 1)knt + a^{1}^{–}^{n} 
x  t  k 0
–k 2.303 k 2
(n – 1)kn 
0  
First 
a 
log_{10} (a – x)  t  log 10 a  
Second  (a – x)^{–}^{1}  t  a 1  
n th  (a – x)^{1}^{–}^{n}  t  a1n 
Plots from integrated rate equations
Plots of rate Vs concentrations [Rate = k(conc.)^{n} ]
 Van’t Haff differential Method : The rate of reaction varies as the n^{th} power of the concentration Where
‘ n‘
is the order of the reaction. Thus for two different initial concentrations C1
and
C2 equation, can be written in
the form,
 dC1= kC n
and
 dC2 = kCn
dt 1 dt 2
Taking logarithms,
log
æ – dC1 ö = log
k + n log C
…..(i)
10 ç ÷ 10
è dt ø
10 1
and
log
æ – dC2 ö = log
k + n log C
…..(ii)
10 ç ÷ 10
è dt ø
10 2
Subtracting equation (ii) from (i),
 dC1 dt
determined.
and
 dC2
dt
…..(iii)
are determined from concentration Vs time graphs and the value of
‘ n‘
can be
 Ostwald’s isolation method (Initial rate method) : This method can be used irrespective of the number of reactants involved g., consider the reaction, n_{1} A + n_{2} B + n_{3}C ® Products .
This method consists in finding the initial rate of the reaction taking known concentrations of the different reactants (A, B, C). Now the concentration of one of the reactants is changed (say that of A) taking the concentrations of other reactants (B and C) same as before. The initial rate of the reaction is determined again. This gives the rate expression with respect to A and hence the order with respect to A. The experiment is repeated by changing the concentrations of B and taking the same concentrations of A and C and finally changing the concentration of C and taking the same concentration of A and B. These will give rate expressions with respect to B and C and hence the orders with respect to B and C respectively. Combining the different rate expressions, the overall rate expression and hence the overall order can be obtained.
Suppose it is observed as follows:
 Keeping the concentrations of B and C constant, if concentration of A is doubled, the rate of reaction
becomes four times. This means that, Rate µ [A]^{2} i.e., order with respect to A is 2
 Keeping the concentrations of A and C constant, if concentration of B is doubled, the rate of reaction is also This means that, Rate µ [B] i.e., order with respect to B is 1
 Keeping the concentrations of A and B constant, if concentration of C is doubled, the rate of reaction remains unaffected. This means that rate is independent of the concentration of C i.e., order with respect to C is Hence the overall rate law expression will be, Rate = k[A]^{2} [B] [C]^{0}
\ Overall order of reaction = 2 + 1 + 0 = 3.
Example : 5 For a reaction, activation energy (Ea ) = 0 and rate constant
 = 2 ´ 10^{6} s ^{–}^{1} at 300K. What is the value of
the rate constant at 310K. [KCET (Med.) 1999)
(a)
3.2 ´ 10 ^{–}^{12} s ^{–}^{1}
(b)
3.2 ´ 10^{6} s ^{–}^{1}
(c)
6.4 ´ 10^{12} s ^{–}^{1}
(d)
6.4 ´ 10 ^{6} s ^{–}^{1}
Solution : (b) When
Ea = 0, the rate constant is independent of temperature so that rate constant (k) = 3.2 ´ 10^{6} sec ^{–}^{1} .
Example : 6 87.5% of a radioactive substance disintegrates in 40 minutes. What is the half life of the substance
 58 min (b) 135.8 min (c) 1358 min (d) None of these
Solution : (a) Determination of k by substituting the respective values.
k = 2.303 log
t
a a – x
= 2.303 log
40
a
a – 0.875a
= 2.303 log
40
a
0.125a
= 2.303 log 8
40
= 0.051 min^{–}^{1}
\ t = 0.693 = 0.693 = 13.58 min
1 / 2 k
0.051
Example : 7 The halflife period of a first order reaction is 100 sec. The rate constant of the reaction is
[MP PMT 1997; MP PET 2001]
(a)
6.93 ´ 10^{–}^{3} sec^{–}^{1}
6.93 ´ 10^{–}^{4} sec^{–}^{1}
0.693 sec^{–}^{1}
69.3 sec^{–}^{1}
Solution : (a)
k = 0.693 =
t1 / 2
0.693
100 sec
= 6.93 ´ 10^{–}^{3} sec^{–}^{1}
Example : 8 The rate constant of a first order reaction is
3 ´ 10^{–}^{6}
per second. If the initial concentration is 0.10mol, the
initial rate of reaction is [AFMC 1999; Pb. PMT 1999, 2000; BHU 1999; AIIMS 1999; Karnataka CET 2000]
(a)
3 ´ 10^{–}^{5} mol s^{–}^{1}
(b)
3 ´ 10^{–}^{6} mol s^{–}^{1}
(c)
3 ´ 10^{–}^{8} mol s^{–}^{1}
(d)
3 ´ 10^{–}^{7} mol s^{–}^{1}
Solution : (d) Given that, rate constant for first order reaction (k) = 3 ´ 10^{–}^{6} per sec and initial concentration (a) = 0.10 mol
we know that initial rate of reaction = k(a) = 3 ´ 10^{–}^{6} ´ 0.10 = 3 ´ 10^{–}^{7} mol sec^{–}^{1}
Example : 9 In a first order reaction the concentration of reactant decreases from
800mol / dm^{3} to
50mol / dm^{3} in
2 ´ 10^{2} sec . The rate constant of reaction in sec^{–}^{1} is [IIT Screening 2003]
(a)
2 ´ 10^{4}
(b)
3.45 ´ 10^{–}^{5}
(c)
1.386 ´ 10^{–}^{2}
(d)
2 ´ 10^{–}^{4}
Solution : (c)
k = 2.303 log
t
a
^{10} (a – x)
; t = 2 ´ 10^{2} sec ,
a = 800mol / dm^{3} ,
(a – x) = 50mol / dm^{3}
k = 2.303 log 2 ´ 10^{2} ^{10}
800 = 1.386 ´ 10^{–}^{2} sec^{–}^{1}
50
Example : 10 The rate of a gaseous reaction is halved when the volume of the vessel is doubled. The order of reaction is (a) 0 (b) 1 (c) 2 (d) 3
R æ a ö ^{n}
Solution : (b)
 R = k a^{n} (ii)
= kç ÷
2 è 2 ø
Dividing (i) by (ii), 2 = 2^{n} . Hence n = 1 .
Example : 11 The first order rate constant for the decomposition of
N2O5
is 6.2 ´ 10^{–}^{4} sec^{–}^{1} . The halflife period for this
decomposition in seconds is [MNR 1991; MP PET 1997; UPSEAT 2000]
(a) 1117.7 (b) 111.7 (c) 223.4 (d) 160.9
Solution : (a)
t1 / 2
= 0.693 =
k
0.693
6.2 ´ 10^{–}^{4}
= 1117.7 sec
Example : 12 A substance
‘ A‘
decomposes by a first order reaction initially with
 = 00mol
and after
200 min, (a – x) = 0.15mol . For this reaction what is the value of k [AIIMS 2001]
(a)
1.29 ´ 10^{–}^{2} min^{–}^{1}
(b)
2.29 ´ 10^{–}^{2} min^{–}^{1}
(c)
3.29 ´ 10^{–}^{2} min^{–}^{1}
(d)
4.40 ´ 10^{–}^{2} min^{–}^{1}
Solution : (a) Given [a] = 2.00mol, t = 200minute and (a – x) = 0.15mol
k = 2.303 log
t
a
^{10} (a – x)
= 2.303 log
200
2.00
^{10} 0.15
= 1.29 ´ 10^{–}^{2} min^{–}^{1}
Example : 13 The halflife for the reaction, N O ⇌ 2NO + 1 O in 24 hrs. at 30^{o} C . Starting with 10g of N O
how
2 5 2 2 2 2 5
many grams of
N2O5
will remain after a period of 96 hours [KCET 1992]
(a)
1.25g
(b)
0.63g
(c)
1.77g
(d)
0.5g
Solution : (b)
k = 0.693 = 0.69 = 2.303 log
1 Or log 10
= 1.2036 or 1 – log(a – x) = 1.2036
t1 / 2 24
96 ^{10} (a – x)
(a – x)
Or log(a – x) = 0.2036; (a – x) = 0.6258g
Example : 14 Thermal decomposition of a compound is of the first order. If 50% of a sample of the compound is decomposed in 120 minutes, how long will it take for 90% of the compound to decompose [Roorkee 1988] (a) 399 min (b) 2.99 min (c) 39.9 min (d) 3.99 min
Solution : (a) Half life of reaction =120 min
k = 0.693 = 0.693 = 5.77 ´ 10^{–}^{3} min ^{–}^{1}
t1 / 2 120
Applying first order reaction equation, t = 2.303 log
k
a ; If a = 100, x = 90
^{10} (a – x)
or (a – x) = 10
So, t = 2.303 . log
5.77 ´ 10^{–}^{3} ^{10}
10 = 2.303 = 399 min
5.77 ´ 10^{–}^{3}
Example: 15 For a reaction A + 2B ® C + D , the following data were obtained [AIIMS 1990]
Expt. Initial concentration Initial Rate of formation of D
(moles litre^{–1}) (moles litre^{–1} min^{–1}) 

S. No.  [A]  [B]  
1.  0.1  0.1  6.0 ´ 10 ^{–}^{3} 
2.  0.3  0.2  7.2 ´ 10 ^{–}^{2} 
3.  0.3  0.4  2.88 ´ 10 ^{–}^{1} 
4.  0.4  0.1  2.4 ´ 10^{–}^{2} 
The correct rate law expression will be
(a)
Rate = k[ A][B]
 Rate = k[ A] [B]^{2}
 Rate = k[ A]^{2}[B]^{2}
 Rate = k[A]^{2}[B]
Solution: (b) From 1 and 4, keeping [B] constant, [A] is made 4 times, rate also becomes 4 times. Hence rate µ [A]. From 2 and 3 keeping [A] constant, [B] is doubled, rate becomes 4 times. Hence rate µ [B]^{2} . Overall rate law will be : rate = k[A][B]^{2} .
Example: 16 The rate of elementary reaction, A ® B , increases by a 100 when the concentration of A is increased ten
folds. The order of the reaction with respect to A is [CPMT 1985]
(a) 1 (b) 10 (c) 2 (d) 100
n n 1
é 1 ù ^{n}
Solution: (c)
R = k[A]
; Also, 100R = k[10 A]
; 100 = êë10 úû
; \ n = 2
“It is the sum of the number of molecules of reactants involved in the balanced chemical equation”.
Or
“It is the minimum number of reacting particles (Molecules, atoms or ions) that collide in a rate determining step to form product or products”.
 Molecularity of a complete reaction has no significance and overall kinetics of the reaction depends upon the rate determining Slowest step is the ratedetermining step. This was proposed by Van’t Hoff.
Example :
NH 4 NO2 ® N 2 + 2H 2 O NO + O3 ® NO2 + O2 2NO + O_{2} ® 2NO_{2}
(Unimolecular) (Bimolecular) (Trimolecular)
 Molecularity of a reaction can’t be Zero, negative or
 Molecularity of a reaction is derived from the mechanism of the given
 Molecularity can not be greater than three because more than three molecules may not mutually collide with each
Mechanism
Step I :
A
2A ® A2
 3B
2 2 2
® A2 B3
(Bimolecular)
Step II :
A + 1 B
2 2 2
® A2 B
(Trimolecular)
Step III :
A2 B + B2
® A2 B3
(Bimolecular)
Example : Decomposition of
H 2 O2
2H 2 O2 ® 2H 2 O + O2 H 2 O2 ® H 2 O + O
H 2 O2 + O ® H 2 O + O2
(Overall reaction Mechanism) (Slow)
(Fast)
Rate =
K[H 2 O2 ] ; The reaction is unimolecular
(1) Pseudo Unimolecular Reaction : Reaction whose actual order is different from that expected using rate
law expression are called pseudoorder reaction. For example,
RCl + H 2 O ® ROH + HCl
Expected rate law :
Rate = k[RCl][H _{2}O] ; Expected order = 1 + 1 = 2
Actual rate law : Rate = k[RCl]; Actual order =1
Because of water is taken in excess amount; therefore, its concentration may be taken constant. The reaction is therefore, pseudo first order. Similarly the acid catalysed hydrolysis of ester, viz.,
RCOOR¢ + H _{2}O ⇌
RCOOH + R¢OH
(follow first order kinetic) : Rate = k[RCOOR¢]
Those reactions which may have order of reaction as one while molecularity of reaction 2 or more than two are as follows :
Examples : (i)
2N 2 O5 ® 4 NO2 + O2 ; Order = 1; molecularity = 2
 CH 3 COOC2 H5
 H O ¾¾^{H}¾+ ® CH COOH + C H OH ; r = k[CH COOC H ]

Order =1 , Molecularity = 2

 inversion of cane sugar : C H O + H O ¾¾^{H}¾+ ® C H O + C H O
Sucrose glucose fructose
Order = 1, Molecularity = 2

 (CH 3) CCl + OH ^{–} ® (CH 3) COH + Cl ^{–}
Order = 1, Molecularity = 2
(v) 2H 2 O2 ¾¾^{P}¾^{t}^{,} ® 2H 2 O + O2
Order = 1, Molecularity = 2
Difference between Molecularity and Order of reaction
Molecularity  Order of Reaction 
It is the number of molecules of reactants terms taking part in  It is sum of the power of the concentration terms 
elementary step of a reaction.  of reactants in the rate law expression. 
Molcularity is a theoretical value  Order of a reaction is an experimental value 
Molecularity can neither be zero nor fractional.  Order of a reaction can be zero, fractional for 
integer.  
Moleculaity has whole number values only i.e., 1, 2, 3, etc.  Order of a reaction may have negative value. 
It is assigned for each step of mechanism separately.  It is assigned for overall reaction. 
It is independent of pressure and temperature.  It depends upon pressure and temperature. 
Some theories, which explain the reaction rate, are as follows:
(1) Collision theory
 The basic requirement for a reaction to occur is that the reacting species must collide with one another. This is the basis of collision theory for
 The number of collisions that takes place per second per unit volume of the reaction mixture is known as collision frequency (Z). The value of
collision frequency is very high of the order of 10^{25} to 10 ^{28} in case of binary collisions.
 Every collision does not bring a chemical change. The collisions that actually produce the product are effective collisions. The effective collisions, which bring chemical change, are few in comparison to the total number of The collisions that do not form a product are ineffective elastic collisions, i.e., molecules just collide and disperse in different directions with different velocities.
 For a collision to be effective, the following two barriers are to be
 Energy barrier : “The minimum amount of energy which the colliding molecules must possess as to make the chemical reaction to occur, is known as threshold energy”.
 In the graph ‘E‘ corresponds to minimum or threshold energy for effective
 There is an energy barrier for each The reacting species must be provided with sufficient energy to cross the energy barrier.
 Orientation barrier : The colliding molecules should also have proper orientation so that the old bonds
may break and new bonds are formed. For example,
NO2 (g) + NO2 (g) ® N 2 O4 (g).
During this reaction, the
products are formed only when the colliding molecules have proper orientation at the time of collisions. These are called effective collisions.
 Thus, the main points of collision theory are as follows,
 For a reaction to occur, there must be collisions between the reacting
 Only a certain fraction of the total number of collisions is effective in forming the
 For effective collisions, the molecules should possess sufficient energy as well as
 The fraction of effective collisions, under ordinary conditions may vary from nearly zero to about one for ordinary Thus, the rate of reaction is proportional to :
 The number of collisions per unit volume per second (Collision frequency, Z) between the reacting species
 The fraction of effective collisions (Properly oriented and possessing sufficient energy), f
i.e., ; Where f is fraction of effective collision and Z is the collision frequency.
 The physical meaning of the activation energy is that it is the minimum relative kinetic energy which the reactant molecules must possess for changing into the products molecules during their This means that the
fraction of successful collision is equal to e ^{–} ^{E}a / RT called Boltzmann factor.
 It may be noted that besides the requirement of sufficient energy, the molecules must be properly
oriented in space also for a collision to be successful. Thus, if
Z AB
is the collision frequency, P is the orientation

factor (Steric factor) then, k = PZ .e – Ea / RT . If we compare this equation with Arrhenius equation. k = Ae – Ea / RT
We know that preexponential form ‘A‘ in Arrhenius equation is,
Concept of activation energy
A = PZ AB .
 The excess energy (Over and above the average energy of the reactants) which must be supplied to the
reactants to undergo chemical reactions is called activation energy (Ea ), Ea = E(Threshold energy) –
E(Reactants)
Activation energy = Threshold energy – Average kinetic energy of the reacting molecules.
 Zero activation energy = Fraction of effective collision (f) will be very large = Very fast reaction (Instantaneous reaction).
 Low activation energies = Fraction of effective collision (f) will be large = Fast
 High activation energies = Fraction of effective collision (f) will be small = Slow
 When the colliding molecules possess the kinetic energy equal to activation energy, the atomic configuration of species formed at this stage is different from the reactants as well as products. This stage is called the activated state or transition state and specific configuration of this state is called activated complex. In other words, we can say that, A collision between high energy molecules overcomes the forces of repulsion and brings the formation of an unstable molecule cluster called the activated complex. The life span of an activated complex is very small. Thus, the activated complex breaks either into reactants again or new substances, e., products.
 The activation energy
(Ea )
depends upon the nature of chemical bonds undergoing rupture and is
independent of enthalpies of reactants and products.
 According to the concept of activation energy, the reactants do not change directly into the The reactant first absorb energy equal to activation energy and form activated complex. At this state, the molecules must have energy at least equal to the threshold energy. This means that the reaction involves some energy barrier which must be overcome before products are formed. The energy barrier is known as activation energy barrier.