This document discusses the concepts of order of a reaction and molecularity of a reaction. It defines order of a reaction as the sum of the exponents of the concentration terms in the rate law equation, while molecularity is the sum of the stoichiometric coefficients of the reactants in the balanced chemical equation. Order of a reaction can be fractional or zero but molecularity is always an integer. The document also provides examples of first order reactions and derives the integrated rate expression for a first order reaction.

1. CHEMISTRY PAPER I STD : XII
Lecture no : 3
aA+bB cC+dD
R  Aa Bb
 Where a&b indicates actual
no of molecules taking part in
the rxn
a+b Molecularity
1) Defn
2) Indicates actual no. of Molecules
taking part in the rxn
3) N2 + 3H2 2NH3
Molecularity =  = 4
S+O2 SO2
Molecularity =  = 2
Theoritical Concept
4) Molecularity Can never be
Zero or fraction always has an
integral Value
5) Unimolecular , bimolecular
Law of mass Action

2. CHEMISTRY PAPER I STD : XII
Lecture no : 3
aA + bB cC+dD
Rate  A  B 
Where & indicate
actual no. of molecules
determining rate of rxn
 +  order
Slowest step is the
Rate determining step
Defn Indicate actual no. Molecules
determining rate of rxn
 Experimental Conept
 Order of a rxn can
be Zero ,fraction or
has an integral value
 Zero order rxn ,first order
rxn,second order rxn, etc.
Rate Law Expression:-
e.g. 2N2O5(g)  4NO2(g) + O2(g)
Rate = k[N2O5]
 Denoted by “n”
N2O5(g)  NO2(8) + NO3(g) (Slow)
N2O5(g) + NO3(g)  3NO2(g) + O2(g)(Fast)
 2N2O5(g)  4NO2(g) + O2(g)

3. CHEMISTRY PAPER I STD : XII
Lecture no : 3
Differential Rate Equation:-
Types of Reaction on basis of order of Reaction
1. 1st order : Rate depends upon concn . of one reactant
Eg : a)
b)
2. 2nd order : Rate depends upon concn . of Two reactant
Eg : a)
b)
3. Zero order : Rate depends upon concn . of Reactant
Eg : a)
b)
4. Fraction order: Rate depends upon concn . of Reactant
Eg : a)
b)
NH4 NO2  N2 + 2 H2O
C12H22O11 + H2O  C6H12O6 + C6H12O6
CH3COOC2H5 + NaOH  CH3COONa + C2H5OH
2 HI  H2 + I2
2 NO + O2  2 NO2
2 NO + Br2  2 NOBr
H2 + Br2  2 HBr
CO + Cl2  COCl2

4. CHEMISTRY PAPER I STD : XII
Lecture no : 3
Differential Rate Equation:-
Unit of K for ‘nth’ order is (mol dm–3)1–n s–1
order value of ‘n’ unit of K
zero
first
second
three
n = 0
n = 1
n = 2
n = 3
mol dm–3 s–1
s–1
mol dm–3 s–1
mol–2 dm6 s–1

5. CHEMISTRY PAPER I STD : XII
Lecture no : 3
Order of Reaction Molecularity of a Reaction
1) It is the sum of the exponents of
the concentration terms in the
rate law equation
1) It is the sum of stoichiometric co-
efficient of the reactants taking
part in the reaction.
2) It need not be a whole number
i.e.,
can be fractional as well as zero.
2) It is always an integral whole number
and never zero.
3) On basis of order of reaction , the
reaction are classified as zero,
first, second order, etc.
3) on basis of molecularity the
reactions are classified as
uni, bi, ter molecular reactions
etc.

6. CHEMISTRY PAPER I STD : XII
Lecture no : 3
Order of Reaction Molecularity of a Reaction
4) It can be determined
experimentally only
It is a theoretical concept.
5) It cannot be obtained from a single
balanced chemical equation.
It can be obtained from a balanced
chemical equation.
6) Order of a reaction can change with
the conditions such as pressure,
temperature, concentration.
Molecularity of reaction does not
change with conditions such as
pressure temperature, concn
7) It help us to understand
mechanism of the reaction
It does not tells us anything about
the mechanism of the reaction.
8) It can be determined for elementary
as well as complex reactions
It can be determined only for
elementary reactions

7. CHEMISTRY PAPER I STD : XII
Lecture no : 3
characteristic of first order reaction.
dC
Rate (r) kC
dt
 1)
dC/dt
C
2) The unit of K is sec-1 or time -1
3) The integrated rate equation is
0
10
t
C2.303
K log
t C

Rearranging above equation we get,
0
10
t
C
log
C
0
10
t
C Kt
log (y mx)
C 2.303
 
t
slope = k/2.303.

8. CHEMISTRY PAPER I STD : XII
Lecture no : 3
5) Further Rearranging integrated
rate equation we get,
10 t 10 0
Kt
log C log C (y mx)
2.303
  
Log Co
Log10Ct
Slope =
k
2.303

Time(t)
(t1/2)
Initial conc. Of reactant
6)
1/2
0.693
t
K


9. CHEMISTRY PAPER I STD : XII
Lecture no : 3
Integrated rate Expression
For 1st order (n=1) chemical rxn:
R=KCR
-dc
dt = KC
-dc
c = Kdt (1)
Integrating eqn (1)
= Ktt
o
-l0gect-10geco
l0geco-l0gect = Kt
l0ge co
ct
 =Kt
=Kt
co
=ekt
Co =Ct ekt
Ct=Co e- kt
2.303 log10
Co
Ct
= kt
K=
2.303
t
Intergrated rate Expression 1st Order
 
ct t
t o
co
dc
K dt
c 
 
   
 ct
e co
log c
=  t
o
K t
o
e
t
C
log
C
 
 
 
o
10
t
C
log
C
 
 
 
Time = 0, concn = Co
Time = t, concn = Ct
C0
Ct

10. CHEMISTRY PAPER I STD : XII
Lecture no : 3
0
10
t
C2.303
K log ...(1)
t C

If time, t = 0 then Let initial concentration of the reactant = C0
& when time = t1/2 , then let the concentration of reactant Left = C0/2
Substituting these value in equation 1
0
10
1/2 0
C2.303
K log
t C /2

10
1/2
2.303
K log 2
t

1/2
2.303 0.3010
K
t


1/2
0.693
t
K
