CHEMICAL KINETICS

c) Very slow reactions : The chemical

SYNOPSIS
reactions which complete in very long time
 Chemical Kinetics is the branch of chemistry are called very slow reactions.
that deals with Ex -1 :-Rusting of Iron in presence of air and
a) Rate of reactions moisture
b) Factors influencing the rate of reaction 4 Fe  3O2  xH 2O  2 Fe2O3 .xH 2O.
c) Reaction mechanism  It is not possible to determine the rates of very
 Thermodynamics tells about only the feasibility fast and very slow reactions by conventional
of a chemical reaction where as chemical methods.But the rates of reactions with moderate
kinetics tells about the velocity of reaction. speed can be determined.
Kinetics studies not only help us to determine
the speed (or) rate of a chemical reaction but Reacton Rate (OR) Rate of Reaction:
also describe the conditions by which the  The decrease in the concentration of the reactant
reaction rates can be altered. per unit time (or) increase in the concentration
Ex:- Thermodynamic data indicates the diamond of the product per unit time is called rate of a
shall convert to graphite but in reality the reaction.
conversion rate is so slow that the change is not  The rate of a reaction measured with respect to
perceptible at all. the decrease in the concentration of the reactants
 Based on the velocity of chemical reactions, the .
reactions are classified into three types.  The rate of the reaction measured with respect
a) Very fast (or) instantaneous reactions: to the increase in the concentration of the
The chemical reactions which are completed products .
within the fraction of seconds are called as very  The rate of a reaction at any particular instant
fast reactions. of time during the course of a reaction is the
Ex:-1) Neutralization between strong acids and rate of change of concentration of a reactant (or)
strong bases. a product at that instant of time.
EX-1) : A  B
NaOH ( aq )  HCl( aq )  NaCl( aq )  H 2O( l )
  A    B d A  d B 
2) Precipitation reactions ravg   or rinst  
t t dt dt
NaCl( aq )  AgNO3( aq )  AgCl( s )   NaNO3( aq )
2) N 2  O 2  2 NO
3) Explosive reactions :
Explosion of T.N.T (Tri nitro toulene)  N2  O2  1  NO
rate =   
b) Moderate reactions : The chemical t t 2 t
reactions which are completed in mesurable time (or)
are called as moderate reactions.   N2   O2    NO
Ex:- 1) Inversion of cane sugar rate=  2  2 
t t t
C12 H 22O11( aq )  H 2O( l )  C6 H12O6 ( aq )  C6 H12O6 ( aq ) 3) pP + qQ  rR + sS
glucose fructose
1 P 1 Q 1 R
2)Combustion of hydrogen (or) coal [under rate =    
normal conditions]. p t q t r t

1 1 S 
H 2 ( g )  O2 ( g )  H 2O( g ) =
2 s t
 4) 5 Br

 aq   BrO3  aq   6 H   aq  
d P
3Br2  aq   3H 2O (1) rinst 
dt
 slope
P
d P 
1   Br    BrO3 


 
dt

Concentration of products
Rate   P2 
5 t t  P 
1   H  1   Br2  1   H 2O  .

P1  t  P P  P1
   rav   2
6 t 3 t 3 t t t2  t1 
5) Hg  l   Cl2  g   HgCl2  s  t1 t2 t time
Rate of reaction =
Fig : Average rate of reaction
  Hg   Cl2    HgCl2 
  
t t t W.E-1:The decomposition of N 2O5 in CCl4 at
6) 2HI  g   H 2  g   I 2  g  318K has been studied by monitoring the
Rate of reaction = concentration of N 2O5 in the solution. Initially
1   HI    H 2    I 2  the concentration of N 2O5 is 2.33 molL1 and
  
2 t t t after 184 minutes, it is reduced to 2.08 molL1 .
 The rate of a reaction varies exponentially with The reaction takes place according to the
time of the reaction.
equation 2 N 2O5  g   4 NO2  g   O2  g  .
 The concentration of the reactants in a reaction
varies exponentially with time. Calculate the average rate of this reaction in
 No reaction takes place with uniform rate terms of minutes? What is the rate of
throughout the course of the reaction. production of NO2 during the period?
 The rates of chemical reactions differ from one Sol: Average rate
another, since the number and the nature of the
1    N 2O5   1   2.08  2.33  molL 
1
bonds are different in the different substances     
(reactants or products or both) 2 t  2 184 min 
 The units for the rate of the reaction is 
 6.79  10 molL / min
4  1

mol. lit 1. sec 1

 when the concentration of gases is expressed in It may be understand that
terms of their partial pressures, then the units of 1   NO2 
Rate   6.79  10 4 molL1 / min
the rate will be atm s 1
4 t
R 0   NO2 
 4  6.79  10 4 molL1 min 1
t
 2.72  10 3 molL1 min 1
 R
Concentration of reactants

rav 
R1  t2  t1  WE-2: N 2  3H 2  2 NH 3 the rate of disappearance
 R 
d R of nitrogen is 0.02molL1s 1 . What is the rate
R2 
t of appearance of ammonia?
d R
rinst    slope
dt d  N 2  1 d  NH 3  1 d  NH 3 
Sol:  ; =0.02
dt dt 2 dt 2 dt
t
d  NH 3 
t1 t2 time  0.04 mol.L1s 1
dt
Fig : Instantaneous rate of reaction The rate of appearance of ammonia = 0.04molL1s1 .
WE-3: A2 B is an ideal gas, which decomposes  Reaction (b) involves breakage of 8 bonds and
formation of eight bonds.
1
according to the equation A2 B  A2  B2 . At
2 H H
start, the initial pressure is 100mm of Hg and | |
after 5 minutes, the pressure is 120mm of Hg . H  C  H  2 O  O  O  C  O  2O H
|
What is the average rate of decomposition of
H
A2 B ? Assume T and V are constant. (4) (4) (4) (4)
Sol: The decomposition reaction of gaseous A2 B is
 Reaction (a) is faster than reaction (b)
1
given as A2 B  A2  B2 b) Concentration of the reactants:Except
2
100 0 0 initial reaction zero order reactions, for all other reactions the
100-2x 2x x final reaction rate depends on the concentration of the
100 -2x + 2x + x = 100 + x = 120mm reactants.
x = 20 mm or 2x = 40 mm  rate  (concentration of the reactants) n or
The decrease in pressure of reactant dc dc
  Cn (or)   K Cn
substance A2 B in 5mm is 40mm. dt dt
The rate of decomposition of ‘n’ may be any simple value including zero.
40  For gaseous reactants.
A2 B   8mm min 1  0.133mms 1
5 rate  (pressure of the reactants)n
Factors affecting the Rate of reaction: dP  dP
  P n or  KP n
a) Nature of the reactants: Reactions dt dt
between ionic substances take place much faster  Chemical reactions occur due to the collisions
than the reactions occuring between covalent among the reacting molecules. Hence greater the
substances.Because in ionic reactions there is number of these molecules in unit volume,
no breaking and making of bonds. greater will be the possibility of their collisions
NaCl aq   AgNO3 aq   AgCl s    NaNO3 aq  and higher will be the rate of reaction.
 The reactions between covalent molecules  Eg:- When zinc pieces are added to dilute HCI ,
involve the breaking (cleavage) and the making chemical reaction takes place slowly liberating
(formation) of covalent bonds.
H  H 2 gas. But the same reaction is rapid by taking
Eg : C2 H5OH l   CH3COOH l  
concentrated HCI .
CH3COOC2 H5 l   H 2O l   From the given below graph it is clear that, the
 Reactions which involve lesser bond rate of reaction gradually decreases with time
rearrangements are rapid at room temperature because of the decrease in the concentration of
than those which involve more bond reacting substances with time.
rearrangements.
a) 2 NO  O2  2 NO2
b) CH 4  2O2  CO2  2 H 2O
 Reaction (a) involves breaking of 6 bonds and rate
formation of six bonds.
O
|| time
2  N  O   O  O  2 N O
(4) (2) (6)
 c) Effect of temperature on the reaction From equations (2) and (3)
Rate :The rate of a reaction increases with Ea E
increase in the temperature. ln K 2  ln K1   a
RT1 RT2
 In most cases a rise of 10 0C in temperature
generally doubles the specific rate of the
K 2 Ea  1 1 
reaction. ln    
 Increasing the temperature of the substance K1 R  T1 T2 
increases the fraction of molecues, which collide
with energies greater than activation energy K2 Ea  1 1 
log    
 Ea  Hence increases the rate of reaction. K1 2.303R  T1 T2 
 The ratio of two specific rates measured at Ea E
temperature that differ by 10o C is called the  slope =   a
2.303R 4.576
Temperature co-efficient (R=gas constant)
Ko
t C 10
 Temperature co-efficient = K 2
o
t C

(Normally t o C  25o C; t o C  10  35o C )

For some reactions the ratio approaches the
value of 3
 Hydrogen and oxygen combine very slowly at
ordinary temperature but rapidly at high
temperature.
 A piece of coal will burn rapidly in air when the H 2 g   I 2 g   2 HI  g 
temperature is raised sufficiently. According to Arrhenius, this reaction can take
 Arrhenius equation for temperature dependence place only when a molecule of hydrogen and a
of a rate constant. molecule of iodine collide to form an unstable
 Ea intermediate. It exists for a very short time and
KA e RT
then break up to form two molecules of
K = Rate constant, hydrogen iodide.
A = Arrhenius frequency constant
Ea = Activation energy, H H I H I
I
R= Gas constant  
T = absolute temperature H I H I H I
 Ea
K  A.e RT Intermediate
Taking ln on both sides
 Ea
ln K   ln A  1 Activated Complex
RT C
at temperature T1 equation 1 is
Potential Energy

 Ea
ln K1   ln A   2 A
RT1 H B
H2  I
at temperature T2 equation 1 is 2HI
Reaction coordinate
 Ea
ln K 2   ln A   3
RT1 Above Graph showing plot of potential energy
( since A is constant for a given reaction) Vs reaction coordinate.
 Let k p denote presence of catalyst and ‘ ka ’
denote absence of catalyst.
 E / RT
k p  Ae p ; ka  Ae  Ea / RT
deviding eq.(1) by eq. (2) we get
kp  Ea  E p  / RT
e =  e E / RT
ka
kp  E 
 anti log  
t Energy of ka  2.303 RT 
Fraction of molecules

This area shows

activation fraction of additional
t  10  molecules which
 Catalyst alters
This area react at (t+10) i) rate of reaction ii) path of reaction
shows fraction iii) activation energy iv) threshold energy
of molecules
reacting at t v) rate constant
 Catalyst does not alter
Kinetic energy
i)  G of reaction
d. Effect of Catalyst : ii) energy of reactants and energy of products
 i) Presence of positive catalyst: The function iii)  H iv) S v) Kc
of a positive catalyst is to lower down the  In some reactions, the rate of the reaction is
activation energy. directly proportional to the concentration of the
The grater the decrease in the activation energy catalyst.
higher will be the rate of reaction Ex: Acid catalysed hydrolysis reactions of esters,
In the presence of a catalyst, the reaction follows the rate is proportional to the concentration of
a path of lower activation energy. the acid catalyst.
In this condition, a large number of reacting  Catalyst increases the rate of reaction by making
molecules are able to cross over the energy an alternate path of low activation energy for
barrier and thus the rate of reaction increases. reactant molecules.
MnO2
Ex:- 2 KClO3   2 KCl  3O2
Reaction path
without catalyst
 Note: Pressure of the gases, volume of the
vessel, state of the substance, p H etc also
influence the rate of some reactions.
Reaction W.E-4:At 27 0 C and 37 0 C , the rates of a reaction
Ea path with
Potential energy Ea catalyst are given 1.6  102 molL1s 1 and
as
Reactants 3.2  10 2 molL1s 1 . Calculate the energy of
The effect of a
catalyst is to lower activation for the given reaction.
the energy of activation
Product
Sol: The ratio of specific rates at two different
Reaction coordinate temperatures are given as,
A catalyst changes the
reaction path ( Positive catalyst ) k2 Ea  T2  T1 
log   
ii) Presence of negative catalyst: A negative k1 2.303R  T1T2 
catalyst increases the activation energy of Substituting the values k1 , k2 , T1 , T2 and molar
reaction by forming a new intermediate of high
gas constant R
energy, i.e., by changing the reaction
Ea  1000 10
mechanism. Due to increases activation energy,, 0.301  
some active molecules become inactive, 2.303  8.314 300  310
therefore, rate of reaction decreases Energy of activation  Ea  53kJmol 1
 i) 2 NO  Cl2  2 NOCl ; r  k  NO  Cl2 
2
WE-5:The temperature coefficient of a reaction
is 2 and the rate of reaction at 250 C is
ii) SO2Cl2  SO2  Cl2 ; r  k  SO2Cl2 
3molL1 min 1 . Calculate the rate at 750 C
iii) 2 HI  g   H 2 g   I 2 g  ; r  k  HI  and
2
Sol: According the Arrhenius, for every 100 C rise in
the temperature, the rate of reaction is doubled.
H 2 g   I 2 g   2 HI  g  ; r  k  H 2  I 2 
Rise in temperature = 750 C  250 C  500 C
 5  100 C 1
iv) H 2O2 aq   H 2O l   O2 g  ; r  k  H 2O2 
The increase in the rate of the reaction 2
 25  32 times v) CH3COOC2 H5 aq  NaOH aq 
The rate of reaction
CH3COONa aq  C2 H5OH aq
750 C  3  32  94molL1 min 1
r = K CH 3COOC2 H 5 1 NaOH 1
Rate Law :The equation which relates the rate vi) CHCl3  Cl2  CCl4  HCl
of the reaction and the concentration of the
Rate = k CHCl3 Cl2 
1/ 2
reactants is known as rate equation or rate law.
Rate law is the expression in which reaction rate vii) CH 3COOC2 H 5  H 2O  CH 3COOH  C2 H 5OH
is given in terms of molar concentration of
Rate = k CH 3COOC2 H 5   H 2O 
1 0
reactants with each term raised to some power,
which may (or) may not be same as the  Rate law equation for reversible reaction
stoichiometric coefficient of the reacting species
H 2  I2 
K
  
1
2HI

in a balanced chemical equation. K 2

 nA + mB  Products
1 d  HI
 k1  H 2  I 2   k 2  HI 
2

rate r  A n B m Rate =

2 dt

rate = K  A  n  B  m Rate of forward Rate of backward

Rate = reaction   reaction 
 Rate equation is obtained experimentally.    

Rate Constant: The rate constant of reaction  Rate law equation involving side reactions.
becomes equal to the rate of the reaction when k1
Th227
the concentration of all the reactants are unity, Ac 227
hence the rate constant is also known as the Fr223
k2
specific reaction rate.
 A + B  Products. Rate of formation of Th 227  k1  Ac 227 
 If the initial concentration of B is taken in large
excess than A, then rate depends on A only. Rate of formation of Fr 223  k 2  Ac 227 
 nA  mB  products Rate   k1  k 2   Ac 227 

rate r  A n B m Characteristics of Rate constant:

rate = K  A n  B  0 (i) Rate constant is a measure of the rate of the
reaction. Greater the value of k, faster is the
 'K' is called rate constant or specific rate or rate reaction. Similarly, smaller the value of k
per unit concentrations of the reactants. indicates slower the reaction.
rate (ii) The value of k depends on the nature of the
K
 reactants  reactants. It is a characterstic constant for a
n

particular reaction at a fixed temeperature.

1 n n 1 1
Units of K  mole L sec Different reactions have different values of k.
(iii) The value of the rate constant of a reaction does a
not depend on the concentrations of the (v) Half-life period  t1/ 2   2k   t1/ 2  a 
reactants. Where a = initial concentration
(iv) The value of rate constant for a particular
(vi) Unit of rate constant : mol lit 1s 1
reaction changes with temeperature.
(v) The units of rate constant depend on the order First order reactrions:
1) A  Products r  k1  A1
of reaction.
Order of the reaction :The sum of the powers
dx
of the concentration terms of reactants in the 2) Equation for rate:  k1 a  x 
rate equation is called order of the reaction. dt
 Order of reaction may be zero (or) fraction (or) Equation for rate constant :
negative (or) a whole number (or) integer 2.303 a
k1  log
 Order of the reaction can be determined by t ax
experimental method only. 3) Units for rate constant: sec-1
 For elementary reactions order can be obtained
4) Half life time : t12  a
0
from stoichiometric equation.
Note: A Chemical reaction which occur in single 0.693
step is called elementary reaction. t 12 
k1
Ex 1:- xA  yB  zC  products
5) Let us consider a typical first order gas phase
R  K  A   B  C 
x y z
reaction A  g   B  g   C  g  .
order  x  y  z Let pi be the initial pressure of A and p t the
total pressure of the reaction mixture at time ‘t’
1 3 2.303 pi
Ex 2:- A  B  products then k  t
log
 i  pt 
2p
2 2
Examples :
Rate = k  A  B 2
1 3
2

1
1 3 a) N2O5 g   N2O4 g   O2 g 
2
Order =   2
2 2 b) SOCl (g) SO(g) Cl (g)
2 2 2 2

Zero order reactions: 1

(i) The reaction rate is independent of the c) H2O2(aq) H2O(l)  O2(g)
2
concentration of the reactants d) Acid Hydrolysis of ester.
(ii) Some examples of zero order reactions are 
CH3COOC2H5  H2Oexcess 
H

eg : (I) H 2  g   Cl2  g  
hv
 2HCl  g 

(II) 2NH3  g  

M
N 2  g   3H 2  g 
0 CH 3COOH  C2 H 5OH
hv

(III) 2HI Au

 H 2  I2 e) NH4 NO2 (aq)  N2 (g)  2H2O(l)
(iii) A  product f) Hydrogenation of ethene is an example of
Rate = k  A   first order reaction.
C2 H 4  g   H 2  g   C2 H 6  g 
Co  C x
k 
(iv)
t t Rate = k C2 H 4 
Where C0  Initial concentration of reactant g) Disintegration of radio active elements
C = Concentration of reactant at ‘t’ time Ra  24 He  86
226
88
222
Rn
x  Concentration of product at ‘t’ time Rate = k[Ra]
Second order reactions : Rate = k  CHCl3  Cl2 1/ 2 order = 1.5
a) 2A  products r  K 2  A
2
e. Reaction between H 2 and D2
b) A+B  products r  K 2  A B  Rate = k .PH 2 .PD2
1/ 2

c) Units for rate constant:

order = 1.5
R ate= K [A ]1 [B ] 0
f. Conversion of para hydrogen to ortho
1 1
(or) t 12  k a
hydrogen at high temperature
d) Half life time : t 12
a
Rate   PH 2 
2 3/ 2
Examples : order = 1.5
1) 2 N 2O  2 N 2  O2 Negative order reactions:
2) 2Cl2O  2Cl2  O2 Conversion of ozone into oxygen.
Rate = K  O 3   O 2 
1
alkaline hydrolysis of ester 2

3) CH3COOC2H5  NaOH  Order with respect to oxygen is –1

CH 3COONa  C2 H 5OH Total Order is 1
4) 2 NO2  2 NO  O2 For first order Growth Kinetics
Ni
5) C 2 H 4  H 2  C 2 H 6 It it used in population growth and bacteria
6) 2 HI  H 2  I 2 multiplication
2.303 ax
Third order reactions: K log
a) 3A  products (or) t a
b) 2A+B  products (or) When ‘a’ is initial population and  a  x  is
c) A+B+C  products
population after time ‘t’
r  k3 A (or)
3
nth Order Reactions:
r = K3[A]2[B]1 (or) r = K3[A]1[B]1[C]1 Units for rate constant :
Units for rate constant : lit2.mole-2.sec-1.
litn-1.mole1-n.sec-1. or (atm)1-n sec-1
1
Half life time : t12  When the order of reaction is n
a2
Examples : 1  2n 1  1 
t1/ 2 
a. 2 NO  O2  2 NO2 K (n  1)  a n 1 
b. 2 NO  Cl2  2 NOCl 1
Half life : t12 
c. 2 FeCl3  SnCl2  2 FeCl2  SnCl4 an1
(aq) (aq) (aq) (aq) Pseudo Unimolecular reactions :
Fractional order reactions: The reactions with molecularity greater than or
a. H 2  Br2  2HBr equal to 2 but order is one are called Pseudo
Rate = k  H2   Br2 1/ 2 Order  1.5 unimolecular or Pseudo first order reactions.
b. CO  Cl2  COCl2 Eg : 1. Hydrolysis of ethyl acetate in acid
medium
Rate = k  CO 2  Cl2 1/ 2 order  2.5 2. Inversion of cane sugar
c. COCl2  CO  Cl2 3. Hydrolysis of Acetic anhydride
Rate = k  COCl2 3/ 2 Order  1.5 Note: For any reaction excess amount of reactant
d. CHCl3  Cl2  CCl4  HCl is taken then order w.r.t that reactant is zero.
 Molecularity of the reactions: Difference Between Molecularity
 The sequence of the elementary steps of a and order of Reaction :
chemical reaction is known as reaction
mechanism.
 The slowest elementary step of the reaction is
called rate determining step or rate limiting step.
 The number of atoms (or) molecules or ions
participating in the slowest step is called
molecularity.
 Molecularity cannot be zero (or) fraction. It is
always a whole number (or) integer.
 Molecularity is obtained from reaction
mechanism.
Eg : NH 4 NO 2  N 2  2H 2 O Unimolecular
2HI  H 2  I 2 Bimolecular
2NO  O 2  2NO 2 trimolecular
 The probability that more than three molecules
can collide and react simultaneously is very
small. Hence, the molecularity greater than three
is not observed. It is, therefore, evident that
complex reactions involving more than three
molecules in the stoichiometric equation must
take place in more than one step. Methods of Determination of order
KClO3  6 FeSO4  3H 2 SO4  of Reaction
KCl  3Fe2  SO4 3  3H 2O 1) Trail and Error method or Integrated
from of rate equation method
This reaction which apparently seems to be of
tenth order but it is actually a second order
reaction. This shows that the reaction takes
place in several steps. Which step controls the r
rate of the overall reaction. or slow reaction that Zero order RP 0

t
is called as rate determining step involves just  
two species.
Consider the decomposition of hydrogen x a  a  x
peroxide which is catalysed by iodide ion in an x  kt (or ) K 
t t
alkaline medium.
I 
First order RP
2H 2O2   2H 2O  O2
Alkalinemedium
2.303 a
K log
The rate equation for this reaction is found to
be
t a  x
 d  H 2O2 
Rate =  k  H 2O2   I  
dt
a
This reaction is first order with respect to both log
a x
H 2O2 and I  . Evidences suggest that this
reaction takes place in two steps. t
 
 4) Ostwald’s Isolation method
This method is useful to determine the order w.r.t
each reactant of a reaction separately by taking
other reactants in excess quantity.
A + B + C  products
Then order with respect to A is n A
Order with respect to B is n B
Order with respect to C is n C
Second order 2R  P overall order of the reaction = n A  n B  n C
1 x
K  WE-6: 75% of a first order reaction is completed
at  a  x 
in 30min. Caluculate (a) half life, (b) rate
constant and (c) time required for 99.9%
completion of the reaction.
x Sol: Time required for 75% completion is 2 half

a  a  x
lifes = 30min.
t
    =15 min
(a) Half-life t 1
2

Second order R1  R2  P
0.693 0.693
b a  x (b) Rate Constant  k     0.046min1
2.303 t1 15
K log
t a  b a b  x  2

(c) Time required for 99.9% of the reaction (t)

2.303 a 2.303 100
t log  log  149 min
log
b a  x  k a  x 0.046 0.1
a b  x
WE-7: A first order reaction is 20% complete in
t
  10min. How long it takes to complete 80%?
Sol: Applying first order integral rate equation,
2) Half-time  t1/ 2  method rate constant K is given as,
n 1
1  t1/2'   a ' '  2.303 a 2.303 100
t1/ 2  n1 ;  ' '    '  k log  log
a  t1/2   a  t ax 10 100  20
where n = order of reaction  0.0223min 1
3) Van’t Hoff Differential method
 log1.25  0.0969 
 dc
 KC n Time required for 80% completion of the first
dt
For two initial concentrations C1 , C2 we have order reaction, t0.8

 dc1  dc2 2.303 100

 KC1n ;  KC2n t0.8  log
dt dt k 100  80
  dc1    dc2  2.303 100
log    log    log  72.2 min
 dt   dt  0.0223 20
n 
 log c1  log c2 
WE-8:The initial concentration of ethyl acetate is Sol: Let the pressure of N 2O5  g  decreases by 2x
0.85molL1 . Following the acid catalysed atm. As two moles of N 2O5 decompose to give
hydrolysis, the conentration of ester after
30min and 60min of the reaction are two moles of N 2O4  g  and one mole of O2  g  ,
respectively 0.8 and 0.754molL1 . Calculate the the pressure of N 2O4  g  increase by 2x atm
rate constant and pseudo rate constant. and that of O2  g  increases by x atm.
Sol: Acid catalysed ester follows pseudo first order
2 N 2O5  g   2 N 2O4  g   O2  g 
kinetics. The rate constant k is given as
Start t  0 0.5 atm 0atm 0atm
2.303 a
k log At timet  0.5  2x  atm 2x atm x atm
t ax
Pt  PN 2O5  PN 2O4  PO2
2.303 0.85
k log  2.020  10 3 min 1  or 
30 0.05   0.5  2 x   2 x  x  0.5  x

k
2.303
log
0.80
 1.997  10 3 min 1 x  pt  0.5
30 0.046
PN2O5  0.5  2 x
The rate constant (k) is the product of pseudo
 0.5  2  Pt  0.5   1.5  2 Pt
first order rate constant  k '
At t  100 s; Pt  0.512 atm
and concentration of water. (concentration of
water = 55.5molL1 ) PN 2O5  1.5  2  0.512  0.476 atm

k  k '  H 2O  But K 
2.303 P 2.303
log i  log
0.5
Substituting the values, t Pf 100 0.476

1.997 103  k ' 55.5 

2.303
 0.0216  4.98  104 s 1
100
Pseudo rate constant = k '
 3.6  10 5 Lmol 1 min 1 Collision Theory of reaction rates or
Kinetic Molecular theory :
WE-9: For a reaction A+2B  products, when B  Collision theory was proposed by Arrhenius and
is taken in excess, then the rate law expression developed by Max Trautz and William Lewis.
and order is  The main postulates of collision theory are
Sol: For the reaction rate law expression is a) Collisions must occur between the molecules
Rate=K[A]1[B]0 of reacting gases for a reaction to occur.
 Order =1 b) All collisions do not lead to the formation of
products. (Only fruitfull collisions leads to
formation of products)
WE-10: The following data were obtained during c) The minimum amount of energy possessed
the first order thermal decomposition of by the colliding molecules to the formation of
N 2O5  g  at constant volume : products or reaction to occur is known as
threshold energy.
2 N 2O5  g   2 N 2O4  g   O2  g  d) The energy possessed by the molecules at
S.NO Time(s) Total Pressure/(atm) STP is known as normal energy or internal
1. 0 0.5 energy.
2. 100 0.512 e) Normal energy possessed by normal
Calculate the rate constant. molecules is always less than threshold energy.
 f) The energy to be gained by the normal Concept of Activation Energy: The
molecules during the collision to convert into difference between the energy barrier (i.e.,
products is known as activation energy or energy threshold energy) ET and the energy of normal
of activation.
g) Activation energy = Threshold energy - molecules ER is called activation energy Ea .
energy of normal colliding molecules. Ea  ET  ER
h) Activation energy increases, the rate of the
reaction decreases.
i) No. of binary collisions per unit time (Z) is
Rate  Z AB e  Ea / RT

8kT
Z   AB
2
n A nB ;

 AB  collision diameter ;
  reduced mass
j) Specific rate k  p. Z e  Ea / RT or k  Ae  Ea / RT
Where P= probability factor
 Activation energy of HI decomposition reaction Threshold energy
52.8K.J/mole.
 For 2NO2  2NO+O2 the activation energy is
111K.J/mole. So decomposition of NO2(g) is
slower than decomposition of HI(g). Potential energy R
 The collisions in which molecules collide with
sufficient kinetic energy (called threshold Energy of colliding H
molecules
energy) and proper orientation, so as to facilitate P
breaking of bonds between reacting species and
Reaction coordinate
formation of new bonds to form products are
Exothermic reaction  H is  ve
called as effective collisions. Where as improper
orientation makes them simply bounce back and
no products are formed. Threshold energy
For example, formation of methanol from
bromomethane.
Energy of
Potential energy activation P

R H
Energy of colliding molecules

Reaction coordinate
Endothermic Re action H is  ve

 The fraction of activated collisions is smaller If activation energy of forward reaction  Eaf 
than the total number of collisions.
 Actual rate of reaction is much smaller than the is less than that of the backward reaction  Eab  ,
rate of the reaction calculated on the basis of the reaction is exothermic.
the normal collisions.
The heat of the reaction, H  Eaf  Eab The free energy change of a photochemical
reaction may not be negative.
Reactions with lower activation energy are fast
In the synthesis of carbohydrates and formation
and with higher activation energy are slow.
of HCl. G is +ve.
For ionic reactions, the energy of activation is
negligibly small and hence they are Quantum Efficiency OR Quantum yield
instantaneous. Number of molecules participating in
For covalent reactions, the energy of activation photochemical reaction with absorption of
is high and the reactions are time consuming. quanta is called Quantum Efficiency. It is
In the presence of a catalyst alters the path, with expressed as
a new path of low activation energy, the time Number of molecules reacting in a given time

required for a covalent reaction is also low. Number of quanta of light absorbed in the same time
Increasing the concentration of reactants Chemiluminescence: Emission of light in a
increases the rate. This is because of the increase
chemical reaction at ordinary temperatures is
in the collision frequency and increase in
called chemiluminescence.
number of reactant molecules crossing the
Fluorescence: The absorption of energy and
energy barrier.
instantaneous reemitting of the energy is called
Collision Frequency (Z) : Total number fluorescence.
of collisions which occur among the reacting Phosphorescence : The continuous glow of
molecules per second per unit volume is called some substances even after the cutting of source
collision frequency. Its value is given by of light is called phosphorescence
z 2  v 2 n 2 Eg. ZnS
v = average velocity Bioluminescence : The phenomenon of
 = molecular diameter in cm chemiluminescence exhibited by certain living
n = number of molecules per cc. organisms is called Bioluminescence
Eg. light emission by fire flies.
 Rate  k   Ze Ea / RT

Transition State theory: According to this ADDITIONAL INFORMATION

theory, the bimolecular reaction between two Expression for the amount left after n
molecules A 2 and B2 passes through the half lives
formation of activated complex which then
 A0
decomposes to yield the product AB, as shown After one half-life, amount left 
below 2
After two half-lives, amount left
A 2  B2   A 2 B2   2AB
1  A0  A0
Pr oduct
Re ac tan ts  Activated Complex 

The constant ‘A’ has unit of time –1 and is    2

2 2 2
constant for a given reaction
After three half-lives , amount left
At very high temperature rate becomes equal to
frequency factor, i.e., k = A. 1  A   A
  20  30
Photochemical reaction :Reactions which 2 2 2
take place by the absorption of radiations of In general, amount left after n half lives
suitable wavelength  A0
Eg : H 2  g   Cl2  g  
 2HCl  g 
light 
2n
Photosynthesis of carbohydrates in plants takes
place in presence of chlorophyll and sunlight Total time
6CO 2  6H 2 O 
light
 C 6 H12 O 6  6O 2
No.of half-lives = t1
2
 Average life of a first order reaction Some typical linear plots for the
Average life of a first order reaction is the time reactions of different orders :
in which the concentration of the reactant is (a) Plots of rate vs concentrations
reduced to 1/e of its original concentration i.e.,
Rate  k  conc.
n

 A   A0  / e . Thus,
1  A0 zero order 1st Order
Tav  log e
k  A0 / e Rate Rate

0.693
1
 log e e 
1 k Conc Conc
k k
But t1
2

t1
 Tav  2
 1.44t 1 2nd Order 3rd Order
0.693 2

OTHER IMPORTANT RELATIONS Rate Rate

Zero reactions:
Conc  Conc 
2 3
x1 x2 3
(i) t  t (ii) t75%  t50% (iii) t100%  2  t50%
1 2 2 (b) Plots from integrated rate equations
First order reactions: 1st Order
Zero Order
10
(i) t75%  2t50% (v) t90%  t50%
3 Conc A log A
(ii) t87.5%  3t50% (vi) t99%  2t90%
Time t  Time t 
(iii) t93.75%  4t50% (vii) t99.9%  3t90%
(iv) t99.9%  10t50% (viii) t99.99%  4t90%
Second order reactions 2nd Order 1 3rd Order
1
(i) t2 / 3  2  t1/ 2 (ii) t3 / 4  3  t1/ 2 A A  2

(iii) t4 / 5  4  t1/ 2 (iv) t99.9%  999  t50%

Time t  Time t 
To determine the order w.r.t . each reactant,
follow the following rules: (c) Plots of half-lives vs concentration
Change done in conc. Of a ( t1/ 2 a1 n )
Order w.r.t
reactant ( Keeping conc. Of Effect on rate
that reactant
other reactants constant )
(i) doubled Doubled 1 zero order 1st Order
(ii) doubled/Trebled/
No effect 0
Quadrupled/ Halved, etc t1/ 2 t1/ 2
(iii) Doubled 4 times 2
(iv) Trebled 9 times 2
(v) Halved Halved 1 Conc Conc
Reduced to ¼
(vi) Halved 2 and so on
th

2nd Order 3rd Order

t1/ 2 t1/ 2

1/ a 1/ a2
 8. A graph is drawn between the concentration
C.U.Q
of the reactants (taken on y-axis) and the time
of reaction (taken on x-axis). The slope of the
RATE OF REACTION tangent drawn to the graph at a point
1. Under a given set of experimental conditions corresponding to reaction time t sec. gives
with increase in the concentration of the 1) rate of the reaction
reactants, the rate of chemical reaction 2) rate constant of the reaction
1) decreases 2) increases 3) rate of reaction at time t
3) remains constant 4) half life period of the reaction
4) first decreases and increases 9. In which of the following cases, does the
2. In a chemical reaction, rate of a chemical reaction go farthest to completion
reaction increases with temperature. The 1) K=102 2) K=10-2 3) K=10 4) K=1
reason is due to 10. At 298 K, 1 atm, among
1) number of collisions between molecules
A) 2 H 2  O2  2 H 2O
increases
2) decreases in activation energy B) H 2  Cl2  2 HCl
3) increase in the number of the molecules with
C) N 2  O2  2 NO
activation energy
4) kinetic energy of reactants increases D) H 2 SO4  KOH  products, correct order of
3. Which of the following reaction is a fast reaction rates is
reaction at laboratory temperature 1) D>A>C>B 2) D<A<B<C
1) reaction between KMnO4 and oxalic acid 3) D>B>A>C 4) D>B=C>A
2) reaction between KMnO4 and mohr's salt 11. chemical kinetics is a branch of physical
3) hydrolysis of ethyl acetate chemistry deals with
4) thermal decomposition of N2O5 1) structure of molecules
4. K represents the rate constant of a reaction 2) heat changes in a reaction
when log K is plotted against 1/T 3) physical changes in a reaction
(T=temperature) the graph obtained is a 4) rate of reactions
1) curve 12. The rate of a reaction
2) a straight line with a constant positive slope 1) increase with increase in temperature
3) a straight line with constant negative slope 2) decrease with increase in temperature
4) a straight line with no slope 3) does not depend on temperature
5. Slowest reaction among the following under 4) does not depend on concentration
identical conditions is 13. The rate of chemical reaction would
1) NaOH  HCl  NaCl  H 2O 1) increase as the reaction proceeds
2) decrease as the reaction proceeds
2) H   OH   H 2O 3) may increase or decrease during the reaction
3) 2 NO  O2  2 NO2 4) remains constant as the reaction proceeds
14. The factor which does not influence the rate
4) CH 4  2O2  CO2  2 H 2O of reaction is
6. In reactions involving gaseous reactants and 1) Nature of reactants
gaseous products the units of rate are 2) Concentration of the reactants
1) Atm 2) Atm-sec 3) Temperature
3) Atm.sec-1 4) Atm2 sec2 4) Molecular mass
7. In the sequence of reaction 15. The rate at which a substance reacts depends
k1 k2 k3
A  B  C  D K3>K2>K1 then upon its
the rate determining step of the reaction is 1) Active mass 2) molecular mass
1) A  B 2) C  D 3) B  C 4) A  D 3) atomic mass 4) equivalent mass
16. The term dc/dt in a rate equation refers to 23. A catalyst
1) concentration of reactants 1) Increases the heat of the reaction
2) change in concentration of reactants or 2) Decreases the heat of the reaction
products with time 3) Does not alter the heat of the reaction
3) velocity constant of the reaction 4) Increases the number of collisions
4) concentration of products 24. For the reaction 2NO2  2NO+O2 which of
17. The rate of chemical reaction depends on the the following is false ?
nature of reactants because 1) The decrease in [NO2] and the increase in
1) The number of bonds broken in the reactant [NO] proceed at the same rate
molecules and the number of bonds formed in 2) The rate of formation of NO is twice the rate
product molecules changes of formation of O2
2) Some of the reactants are solids at the room 3) The average rates of increase in the
temperature concentration of NO and O2 are expressed as
3) Some of the reactants are coloured d
4) Some of reactants are liquids at room  NO  and d O 2 
dt dt
temperature
d  NO  2 d  O2 
18. The relation between the rate of a simple 4) 
reaction and the concentration 'c' of the dt dt
25. The rate constant is given by the equation
reacting species is given as
1) rate  c K  A.e  Ea / RT which factor should register a
decrease for the reaction to proceed more
1 1 rapidly
2) ra te  3) rate  n
c c 1) T 2) K 3) A 4) Ea
26. Arrhenius equation may be written as
4) rate c n (n=order of reaction)
d ln k Ea d ln k E
19. Dimensions of rate of reaction involves 1)  2)  a2
1) concentration only dT RT dT RT
2) time only d ln k E d ln k E
3)  a 4)   a2
3) both concentration and time dT RT dT RT
4) neither time nor concentration 27. In Arrhenius plot, intercept is equal to
20. Which of the following about the rate
constant k of a reaction wrong ? Ea
1)  2) ln A 3) ln K 4) log10 a
1) it remains unchanged throughout the course R
of reaction
2) it provides a convenient measure of reaction ORDER OF REACTION
rate 28. In the reaction A  B if the concentration of
3 ) i t i s e x p r e
-1
for all
s s e
A is increased by four times, the rate of the
d i n t h e s a m e u n i t ( s e c )

reactions reaction becomes doubled, the order of the

4) the more rapid the reaction, the larger is the reaction is
value of k, the slower the reaction the smaller is 1) Zero 2) 1 3) 1/2 4) 2
its value 29. In the reaction A+B  Products, if B is taken
21. The value of the rate constant of a reaction in excess, then it is an example of
depends on 1) Second order reaction
1) time 2) activation energy 2) Zero order reaction
3) temperature 4) half-life value 3) Fractional order reaction
22. For an irreversible chemical reaction, the 4) First order reaction
concentration of the products with time 30. The unit of rate constant for a second order
1) increases 2) decreases reaction is
3) does not change 1) litre sec. 2) lit.mole sec.
 
3) Moles .Lit.sec 4) moles.sec
1 1
4) some more data required
31. A chemical reaction A+2B  AB2 follows in 39. Units for the rate constant of first order
two steps reaction is
A+B  AB(slow) 1) Sec-1 2) moles.lit-1
AB+B  AB2(fast) 3) lit.mole-1 4) moles3.lit-1.sec-1
Then the order of the reaction is 40. A zero order reaction is one whose rate is
1) 3 2) 2 3) 1 4) 0 independent of
32. The graph drawn between the reaction time 1) temperature of the reaction
and which of the following concentration 2) the concentration of the reactants
term gives a straight line plot passing through 3) the concentration of the products
origin for the first order reaction 4) activation energy
1 41. The dimensions of rate constant of a second
1) log x 2)
a  x  order reaction involves.
1) neither time nor concentration
a 1 2) only time
3) log
ax
4)
a  x 2 3) time and concentration
33. The reaction that obeys the expression 4) time square and concentration
1 42. The decomposition of H2O2 is represented as
t1  H2O2  H2O+O(slow)
ka the order of reaction
2
(O)+(O)  O2(fast)
1) 0 2) 1 3) 2 4) 3 Then the order of the reaction is
34. The rate equation for the hydrolysis of an 1) 1 2) 2 3) 0 4) 3
ester in presence of NaOH is, rate=k[ester]
[NaOH]. If the concentration of NaOH is d  NH 3 
43. represents
increased by 100 times than that of ester, the dt
order of the reaction will be 1) Rate of formation of Ammonia
1) 1 2) 2 3) 0 4) 3 2) Rate of decomposition of Ammonia
35. When molecules of type A react with
3) Rate of consumption of N 2
molecules of type B in one-step process to give
AB2,the rate law is 4) Rate of consumption of H 2
1) rate =K[A]1 [B]2 2) rate=K[A]2 [B]1 44. Which of the following is not a first order
3) rate=K[2A] [B] 4) rate=K[A] [B] reaction
36. The rate expression for a chemical reaction 1) Hydrolysis of an ester in acidic medium
2NO2F  2NO2+F2 is given by rate =K[NO2F] 2) Decomposition of N2O5
The rate determining step may be
1) 2NOF2  2NO2+F2 2) NO2F +F  NO2+F2 3) Decomposition of H 2O2
3) NO2F  NO2+F 4) NO2+F  NO2F 4) Oxidation of nitric oxide
37. The units of rate of reaction and rate constant 45. The order of a reaction
are identical for a 1) can never be zero
1) fraction-order reaction 2) can never be fraction
2) zero-order reaction 3) must be a whole number
3) first-order reaction 4) can be an integer or a fraction or zero
4) second-order reaction 46. The order of a reaction can be predicted with
38. If a reaction obeys the following equation the help of
2.303 a 1) molecularity of the reaction
k log
t a  x  then the order is 2) activation energy of the reaction
3) rate equation of the reaction
1) 0 2) 1 3) 2 4) 3
4) reaction rate
47. For the reaction A  B the rate law is, 54. The time for half change for a zero order
rate=k[A]. which of the following statement reaction is..................
is incorrect ? 1) proportional to the initial concentration
1) The reaction follows first order kinetics 2) proportional to the square root of the initial
2) The t 1 of reaction depends upon initial concentration
2 3) independent of initial concentration
concentration of reactants 4) inversely proportional to the initial conc.
3) k is constant for the reaction at a constant 55. Rate equation for a second order is
temperature 2.303 a 1 a
4) The rate law provides a simple way of 1) K  log 2) K  log
t ax t ax
predicting the concentration of reactants and
1 X 1 a
products at any time after the start of the reaction 3) K  t a(a  x) 4) K  t 2 (a  x)
48. The decomposition of Cl2O is
56. The hydrolysis of ethyl acetate
1) explosive reactions 2) second order reactions 
H
3) first order reactions 4) thermal reactions CH 3COOC2 H 5  H 2 O  CH 3COOH C2 H 5OH
49. CO=initial concentration of the reactant is a reaction of
Ct=concentration of the reactant at time t, 1) Pseudo first order 2) Second order
k=rate constant of the reaction. Then the 3) Third order 4) Zero order
equation applicable for a first order reaction 57. Radioactive decay follows which order kinetics?
is 1) zero 2) 1 3) 2 4) 3
1) Ct=C0e -kt
2) Ct=C0e kt 58. A reaction involving two different reactants
C0 can never be
3) C0=Cte -kt
4)  1 1) Second order reaction 2) First order reaction
Ct 3) Unimolecular reaction 4) Bimolecular reaction
50. In a first order reaction fraction of the total 59. What is the order of a reaction which has a
concentration of the reactant varies with time rate expression?
't' is equal to
rate  k  A  B 
3/ 2 1

1
1) e kt 2) 100.434kt 3) n 4) e kt 3 1
2
1) 2) 3) zero 4) None
51. For a first order reaction, if ‘a’ is the initial 2 2
concentration of reactant, then the half life
time is
MOLECULARITY
1) independent of a 2)  a 60. The molecularity of a reaction will be
3)  a2 4)  a3 1) fractional 2) zero
52. The rate expression for a reaction is 3) positive whole number 4) negative
61. Which of the following is wrong
dx 1 3
1) order of the reaction is negative, positive or
 K [ A] [ B]
2 2 , the overall order of the
dt fractional
reaction is. 2) molecularity of the reactions is always equal
3 to the sum of stoichiometric co-efficients
1) 2 2) 12 3) 2 4) 1 3) the order of a reactions may be zero
53. Which of the following statements is false? 4) half life is independent of the concentration
1)a fast reaction has a large rate constant and of reactants in first order reaction
short half-life 62. Which statement is correct ?
2) Half life depends on concentration of 1) Molecularity of a reaction is same as the order
reactants for first order reaction. of reaction
3)For a first order reaction,the half-life is 2) In some cases order of reaction may be same
independent of concentration as the molecularity of the reaction
4)The half-life of a reaction is half the time 3) Molecularity may be zero
required for the reaction to go to completion 4) Molecularity may be fractional
63. Which of the following cannot be 72. The excess of energy required for the reactant
determined experimentally. molecules to undergo a reaction is
1) Order 2) Rate 1) Potential energy 2) Kinetic energy
3) Rate constnat 4) Molecularity 3) Thermal energy 4) Activation energy
64. Which of the following statements regarding 73. Threshold energy (TE), internal energy of
molecularity of the reaction is correct? reactants (IE) and energy of activation (AE)
1) Molecularity relates to mechanism of reaction are related as
2) It cannot be negative or fractional 1) AE = TE + IE 2) TE = AE + IE
3) Molecularity of a complex reaction has two 3) IE = AE - TE 4) TE = AE = IE
(or) more steps and each individual step has its 74. The energy to be possessed by the molecule
own molecularity.
participating in the reaction to give the
4) All are correct
products
COLLISION THEORY 1) < activation energy 2) threshold energy
65. To increase the rate of a chemical reaction, 3) < average energy
catalyst 4) threshold energy + average energy
1) increase the activation energy 75. For a given reaction which one is higher than
2) decrease activation energy the rest among the following
3) reacts with products 1) Average energy 2) Threshold energy
4) do not changes the activation energy 3) activation energy 4) Normal energy
66. The energy of activation of a reaction is 76. The value of energy of activation for radio
dependent on active decay is
1) temperature 2) pressure 1) high 2) low 3)zero 4) moderate
3) concentration 4) nature of reactants
77. In arrhenius equation, the fraction of
67. If the activation energy of both the forward
effective collisions is given by
and the backward reactions are equal , H of
1) K=Ae-Ea/RT 2) A 3) e-Ea/RT 4) RT
the reaction is
78. On increasing the temperature by 100C,
1) zero 2) +Ve 3) -Ve
4)cannot be predicted 1) number of collisions get doubled
68. For the exothermic reaction A+B  C+D. 2) value of rate constant does not change
3) energy of activation increases
H is the heat of reaction and E a is the
activation energy. The activation energy for 4) specific rate of the reaction gets doubled
the formation of A+B will be 79. The threshold energy of a chemical reaction
1) Ea 2) H 3) Ea+ H 4) H - Ea depends upon
69. The rate constant (K1) of one reaction is found 1) nature of reacting species 2) temperature
to be double that of the rate constant of (K2) 3) concentration of species
another reaction. Then the relationship 4) number of collisions
between the corresponding activation 80. Activation energy is _____ to rate of reaction
energies of two reactions (E1 and E2) can be 1) directly proportional 2)inversely proportional
represented. 3) equal 4) not related
1) E1 > E2 2) E1 < E2 3) E1 = E2 4) E1 = 4E2 81. The rate of a reaction can be increased in
70. Collision theory is applicable to general by all the factors except
1) Unimolecular reactions 1) using a catalyst
2) Bimolecular reactions 2) increasing the temperature
3) Trimolecular reactions 3) increasing the activation energy
4) Tetra molecular reactions 4) increasing the concentration of reactants
71. The rate constant is given by the equation 82. The energy of activation of positive catalyzed
K  P.Ze  Ea / RT Which factor should register reaction as compared to that of an
a decrease for the reaction to proceed more uncatalyzed reaction is
rapidly? 1) more 2) less
1) T 2) Z 3) Ea 4) P 3) same 4) may be more or less
83. For producing the effective collisions, the 10. Ionic reactions are faster then covalent reactions
colliding molecules must posses 12. Rate of a reaction increases with increase in
1) a certain minimum amount of energy temperature
2) energy equal to greater than threshold energy 13. As the reaction proceeds, the rate of the reaction
3) proper geometry decreases.
4) threshold energy and proper orientation 14. Molecular mass does not influence the rate of
the reaction where as others influence
15. Rate of the reaction is directly proportional to
C.U.Q - KEY
the product of active masses of the reactants
1) 2 2) 3 3) 2 4) 3 5) 4 6) 3 7) 1
19. mol1 n .lit n 1.sec 1
8) 3 9) 1 10) 3 11) 4 12) 1 13) 2 14) 4 21. Rate constant does not depend an activation
15) 1 16) 2 17) 1 18) 4 19) 3 20) 3 21) 3 energy
22) 1 23) 3 24) 3 25) 4 26) 4 27) 2 28) 3 22. The concentration of the reactants decreases
while that of products increases with time.
29) 4 30) 3 31) 2 32) 3 33) 3 34) 1 35) 1
1 d  NO2  1 d  NO d O2 
36) 3 37) 2 38) 2 39) 1 40) 2 41) 3 42) 1 24. rate    
2 dt 2 dt dt
43) 2 44) 4 45) 4 46) 3 47) 2 48) 2 49) 1 25. with decrease in the activation energy, the rate
50) 4 51) 1 52) 1 53) 2 54) 1 55) 3 56)1 of the reaction increases
57) 2 58) 3 59) 2 60) 3 61) 2 62) 2 63) 4 26. Another form Arrheniou
64) 4 65) 2 66) 4 67) 1 68) 3 69) 2 70) 2 Ea  1 
27. ln K     ln A ; y  mx  c
71) 3 72) 4 73) 2 74) 2 75) 2 76) 3 77) 3 R T 
78) 4 79) 1 80) 2 81) 3 82) 2 83) 4    
28. r  k A n  (1); 2r  K 4 A n (2)

C.U.Q - HINTS From eq (1) & (2) , we get n  1/ 2

1. With increase in the concentration of the 29. For any reaction excess amount of reactant is
reactants, the rate of the chemical rections taken then order w.r.t that reactant is zero.
increases due to more number of effective 30. mol1nlit n1 sec1 for n = 2 , if becomes
collision. mol 1lit 1 sec 1 .
2. Increasing the temperature of the substance 31. Order of the reaction can be known from the
increases the fraction of molecues, which collide no. of reactants involved in rate determining step
with energies greater than activation energy (slow step)
 Ea  Hence increases the rate of reaction.
3. Due to lesser number of bond rearrangements
4.

log k
32.

1
T 1
33. t1/2  for n = 2, the expression becomes
5. Due to more number of bond rerrrangements Ka n 1
6. For the gaseous reactants units of rate are
1
Atm.sec 1 t1/2 
Ka
7. Slowest step is the rate determining step for
34. If the concentration of one of the reactants is
which ' K1 ' is least. much higher than the other reactant, the reaction
9. Higher the rate constant. more will be the extent becomes an example of first order.
of completion
35. A  2 B  AB2 51. For first order reaction , t1/2 does not depend as
36. The No. of species involved in rate 0.693
determination step is equal to stoichiometric the initial concentration . t1/2 
K
coefficient of concentration terms in rate 1 3
equation. 52. order    2
2 2
37. Unit of rate constant for zero order reaction and
unit of rate of the reaction are same 53. For first order reactions, t1/2 is independent of
initial concentration
mol.lit 1.sec 1
1 a
38. Rate expression for first order is K  2.303 log a 54. t1/2  n 1 for n = 0  t1/2 
t ax ka k
1 n n 1 1
39. mol .lit .sec for first order n = 1 56. Water is present in excess. Therefore, the rate
Hence the unit becomes sec 1 of reaction is independent of conc. of H 2O
40. For zero order reactions, the rate of the reaction 3 1
is independent of the concentration of the 59. Order = 2  1  2
reactants. 60. Molecularity of the reaction is always positive
41. Expect for first ofder reactions, the unit of rate whole number but can not be zero, fractional or
constant depends as the dimension negative.
 mol .lit .sec  concentration and time 63. Collision
1 n n 1 1

reactions.
theory is applicable for bimolecular

42. Decomposition of H 2O2 is example of first 65. A catalyst increases the rate of the reaction by
order reaction. decreasing the activation energy
66. Activation energy depends on the nature of the
43. 2 NH 3  N 2  3H 2 reactants
1 d  NH 3   Ea back wardreaction   Ea  forward reaction
rate of decomposition of ammonia   68. H
2 dt
for exothermic reaction .
44. 2 NO  O2  2 NO2 ia an example of second 69. Higher the rate constant, lesser the activation
order reaction whereas the remaining given are energy
first order reactions. 71. Ea decreases
45. Conceptual.
73. Threshold energy = Activation energy + Internal
46. Order of the reaction can be known by the rate
energy
equation of the reaction 75. Threshold energy = Activation energy + Internal
47. For first order reaction , t1/2 does not depend asenergy
0.693 76. Because H  0
the initial concentration . t1/2  78. For every 100 C rise in temperature, the rate of
K
the reaction generally gets doubled
48. Decomposition of Cl2O is an example of second 79. Threshold and activation energies depend on the
order reaction. nature of the reactants
c0 78. For every 100 C rise in temperature, the rate of
49. kt  2.303  log c the reaction generally gets doubled
t
79. Threshold and activation energies depend on the
c0 ct nature of the reactants
kt  ln ,   kt  ln
ct c0 80. Lesser the activation energy, more will be the
rate of the reaction.
 kt ct
e  ,  ct  c0 e  kt
81. Activation energy is inversely proportional to
c0 rate of the reaction.
82. A positive catalyst increases the rate of the
50.  ct  c0 e  kt , ct  e  kt reaction by decreasing the activation energy
 6. Observe the following reaction
LEVEL-I (C.W) A g   3B g   2C g 

 d  A 
RATE OF REACTION & FACTORS The rate of this reaction    is
1. The rate of gaseous reaction is given by  dt 
K[A] [B]. If the volume of reaction vessel is
3×10-3 mole lit -1 min -1 . What is the value of
1
reduced to
4
of initial volume the reaction d B 
- in mole lit -1 min -1 ?
rate relative to the original rate is dt
1 1 1) 3  103 2) 9  103
1) 2) 3) 8 4) 16
16 8 3) 103 4) 1.5  10 3
2. The rate of reaction for A  products is 10 7. For which of the following reactions K310 / K300
mole.lit -1 .min -1 when t1=2min. The rate of would be maximum
reaction when t2=12min. in the same units is 1) A  B  C ; Ea  50 kJ
1) >10 2) <10 3) 10 4) 12
2) X  Y  Z ; Ea  40 kJ
3. C12 H 22O11  H 2O  C6 H12O6  C6 H12O6
(excess) ( glu cos e) ( fructose) 3) P  Q  R ; Ea  60 kJ
Rate law is expressed as 4) E  F  G ; Ea  100 kJ
1) r  K C12 H 22O11  H 2O 8. The slope in the activation energy curve is
2) r  K  C12 H 22O11  5.42  103 . The value of the activation energy
is approximately
3) r  K  H 2O 
1) 104 J mol1 2) 104 MJ mol1
4) r  K  C12 H 22 O11  H 2 O 
2
3) 104 KJ mol1 4) 104 J mol1 K 1
4. A chemical reaction was carried at 300K and
9. For the reaction 4NH3  5O 2  4NO  6H 2O,
280K. The rate constants were found to be
K1 and K2 respectively. then the rate of reaction with respect to NH3 is

1) K2  4K1 2) K 2  2 K1 2  103 Ms 1. Then the rate of the reaction

with respect to oxygen in Ms 1
3) K 2  0.25K1 4) K 2  0.5K1
5. The differential rate law for the reaction 1) 2  103 2) 1.5  10 3

H 2  I 2  2 HI is 3) 2.5 103 4) 3  103

10. The rate of formation of SO3 in the reaction
d  H 2  d  I 2   d  HI 
1.   2SO 2  O 2  2SO3 is 100g min 1 . Hence rate
dt dt dt
d H2  d  I2  d  HI  of disappearance of O2 is
2.  
dt dt dt 1) 50g min 1 2) 100g min 1
1 d  H 2  d  I 2  d  HI  3) 20 g min 1 4) 40 g min 1
3.  
2 dt dt dt 11. The rate of the reaction at 400C is 5 units,
then the rate of same reaction at 80 0C is
d H2  d  I2  d  HI 
4. 2  2  (nearly)
dt dt dt 1) 10 units 2) 40 units 3) 20 units 4) 80 units
12. Consider the following reaction 17. Which one of the following statement for
N 2 ( g )  3H 2 ( g )  2 NH 3 ( g ) order of reaction is not correct?
1) Order can be determined experimentally
The rate of this reaction in terms of N2 at T(k)
is 2) Order of reaction is equal to sum of the
powers of concentration terms in differential rate
d  N 2  law
 0.02 m ole .lit  1 .sec  1
dt 3) It is not affected with stoichiometric
 d H 2  coefficient of the reactants
what is the value of (in units of mole 4) Order can not be fractional
dt
18. Which of the following relation is correct for
lit-1 sec-1) at the same temperature.
a first order reaction?
1) 0.02 2) 50 3) 0.06 4) 0.04
13. Rate Equation Units of K (K=rate constant; r=rate of reaction;
I) rate = K[A] a) mole lit-1 sec-1 c = conc. of reactant)
II) rate = K[A] [B] b) lit2 mole-2 sec-1 1) K  r  c2 2) K  r  c
III) rate = K[A] [B] 2
c) sec -1
c r
IV) rate = K d) lit mole-1 sec-1 3) K  4) K 
r c
The correct matching is
1) I - d, II - c, III - a, IV - b 19. The rate cosntant of a reation is
2) I - c, II - d, III - b, IV - a 175 lit 2 mol 2 sec1 . What is the order of
3) I - a, II - b, III - c, IV - d reaction
4) I - b, II - a, III - d, IV - c 1) first 2) second 3) third 4) zero
14.Assertion (A) : Rate of reaction will be 20. If the rate of reaction A  B triples on
doubled, when temperature increased from increasing the concentration of A by 9 times,
298 k to 308 k. then the order of reaction is
Reason (R) : The activation energy of 1) 2 2) 1 3) 1/2 4) 4
reaction decreases with increase in 21. Assertion (A) : For zero order reaction the
temperature. rate of reaction does not decrease with time
1) Both (A) and (R) are true (R) is the correct
Reason (R) : For zero order reaction amount
explanation to (A)
of substance reacted is proportional to time
2) Both (A) and (R) are true but (R) is not the
1) Both (A) and (R) are true (R) is the correct
correct explanation to (A)
explanation to (A)
3) (A)is true but (R) is false
2) Both (A) and (R) are true but (R) is not the
4) Both (A) and (R) are false
correct explanation to (A)
ORDER OF REACTION 3) (A) is true but (R) is false
15. For a reaction 2A+3B  Products, the rate 4) Both (A) and (R) are false
law expression is given by rate = K(A)1  B  . 22. A substance initial concentration (a) reacts
2

according to zero order kinetics. What will

The order of the reaction with respect to A,B be the time for the reaction to go to
and over all order of reaction are completion
1)1,2,1 2)3,2,1 3)1,2,3 4)2,1,3
16. The rate of a certain reaction at different a K a 2K
1) 2) 3) 4) .
times is as follows K a 2K a
Time 0 10 20 30 23. The conversion of A to B follows second order

Rate 3.2  10 3.18102 3.22 10 3.19 102
2 2
kinetics. Doubling the concentration of A will
The order of the reaction is increase the rate of formation of B by a factor
1) 1 2) zero 3) 2 4) 3 1) 4 2) 2 3) 1/4 4) 1/2
24. The rate constant is numerically same for 31. Assertion (A) : Molecularity of a reaction
three reactions of first, second and third or- cannot be more than three
der respectively. Which one is true for rate Reason (R) : Probability of simultaneous
of three reactions, if concentration of reac- collision between more than three particles
tant is greater than 1 M: is very less
1) r1  r2  r3 2) r1  r2  r3 1) Both (A) and (R) are true (R) is the correct
explanation to (A)
3) r1  r2  r3 4) All of these 2) Both (A) and (R) are true but (R) is not the
25. For the chemical reaction A  products, it correct explanation to (A)
is found that the rate increases by a factor of 3) (A) is true but (R) is false
6.25, when the concentration of A is increased 4) Both (A) and (R) are false
by a factor of 2.5. The order of this reaction 32. Assertion (A) : The molecularity of a reaction
with respect to A is: is a whole number other than zero, but
1) 2.5 2) 2 3) 1 4) 0.5 generally less than 3
26. The initial rates for gaseous reaction Reason (R) : The order of a reaction is always
A  3B  AB3 are given below whole number
1) Both (A) and (R) are true (R) is the correct
 A   M   B  M  Rate  M sec1  explanation to (A)
0.1 0.1 0.002 2) Both (A) and (R) are true but (R) is not the
0.2 0.1 0.002 correct explanation to (A)
0.3 0.2 0.008 3) (A) is true but (R) is false
0.4 0.3 0.018 4) Both (A) and (R) are false
order of the reaction is 33. The molecularity of an elementary reaction
1) zero 2) three 3) one 4) two X  2Y  Products is
27. 3/4 th of first order reaction was completed 1) 1 2) 2 3) 3 4) 0
in 32min,15/16 the part will be completed in HALF - LIFE
1) 24 min 2) 64 min 3) 16 min 4) 32 min 34. Which order reaction obeys the expression
28. Initial concentration of the reactant is 1
t1/ 2  in chemical kinetics
1.0M.The concentration becomes 0.9M, 0.8M K .a
and 0.7M in 2 hours,4hours and 6hours 1
respectively ,then the order of reaction is 1) 0 2) 1 3) 2 4) 1
2
1) 2 2) 1 3) zero 4) 3 35. Half life of a zero order reaction is 250sec.
MOLECULARITY t 75% , t100% of the reaction respectively in sec.
29. Molecularity of the following elementary are
reaction is 2NO  O 2  2NO 2 1) 500, 375 2) 375, 500
3) 300, 575 4) 575, 300
1) 0.5 2) 1 3) 2 4) 3
36. The half life period of a first order chemical
30. For which of the following reactions the
reaction is 6.93 minutes. The time required
molecularity and order of the reaction are
for the completion of 99% of the chemical
respectively two and two
1) Ester hydrolysis in acid medium reaction will be (log2=0.3010) ( AIE - 2009 )
2) Inversion of cane sugar in acid aqueous 1) 23.03 minutes 2) 46.06 minutes
solution 3) 460.6 minutes 4) 230.3 minutes
3) Hydrolysis of ethyl acetate in caustic soda 37. 75% of a first order reaction is completed in
aqueous solution. 32 minutes. 50% of the reaction would have
4) Decomposition hydrogen peroxide in acid been completed in
solution 1) 24 mins 2) 16 mins 3) 18 mins 4) 23 mins
38. The half life periods of four reactions labelled 43. The energy profile diagrams of two reactions
by A,B,C & D are 30sec,4.8 min,180sec and are shown in the figure. Then
16 min, respectively. The fastest reaction is
1) A 2) B 3) C 4) D

energy
39. Half-life periods for a reaction at initial
concentrations of 0.1M and0.01 are 5 and 50
minutes respectively. Then the order of A C
reaction is
Reaction coordinates B D
1) zero 2) 1 3) 2 4) 3
40. Assertion (A) : Half life period is inversely 1) Reaction A  B is faster and more exothermic
proportional to rate constant in second order than reaction C  D
reaction 2) Reaction C  D is faster than reaction A  B
Reason (R) : Half life period is always but less exothermic
independent of initial concentration. 3)Reaction C  D is faster and more exothermic
1) Both (A) and (R) are true (R) is the correct than the reaction A  B.
explanation to (A) 4)Reaction C  D 2 1 2 times faster than reaction
2) Both (A) and (R) are true but (R) is not the
correct explanation to (A) A  B at the same temperature
3) (A) is true but (R) is false 44. For a reversible reaction A  B , which one
4) Both (A) and (R) are false of the following statements is wrong from the
given energy profile diagram
COLLISION - THEORY
41. In the equilibrium A  B  C  D , the
activation energy for forward reaction is
25k.cal/mole and that of backward reaction
E B
is 15k.cal/mole. Which one of the following A
statement is correct
1) It is an exothermic process
Reaction coordinate
2) It is an endothermic process
3) It is reaction for which H  0 1) Activation energy of forward reaction is
4) It is a sublimation process greater than backward reaction
42. For an exothermic chemical process, occuring 2) The forward reaction is endothermic
in two steps as 3) The threshold energy is less than that of
i) A  B  X  slow  ii) X  AB  fast  activation energy
4)The energy of activation of forward reaction
The progress of the reaction can be best is equal to the sum of heat of reaction and the
described by
energy of activation of backward reaction.
x 45. Assertion (A) : A catalyst increases the rate
of the reaction
x
Reason(R) : A catalysed reaction proceeds
through a new path having higher activation
1) A B 2) A B energy.
AB AB 1) Both (A) and (R) are true (R) is the correct
explanation to (A)
2) Both (A) and (R) are true but (R) is not the
correct explanation to (A)
x 3) (A) is true but (R) is false
4) Both (A) and (R) are false

3) A B 4) All correct
AB
 LEVEL -I (C.W) - KEY 13. Unit of K  mole1nlit n 1 sec1
1) 4 2) 2 3) 2 4) 3 5) 4 6) 2 7) 4 14. For every 100 C rising in temperature, rate of
8) 1 9) 1 10) 3 11) 4 12) 3 13) 2 14) 3 reaction becomes double due to double the no.
of effective collisions
15) 3 16) 2 17) 4 18)4 19) 3 20) 3 21) 2
15. Order w.r.to A = 1
22) 1 23) 1 24)3 25) 2 26) 4 27) 2 28)3
Order w.r.to B = 2
29) 4 30)3 31) 1 32) 3 33) 3 34) 3 35)2
Over all order = 1 + 2
36)2 37) 2 38) 1 39) 3 40) 3 41) 2 42)2 16. For zero order, rate remains constant with time.
43) 3 44) 3 45) 3 17. Order can be fractional
18. r  K  conc 
n

LEVEL-I - HINTS
19. From units of rate constant, the reaction is third
1. r  K  A B order
 A  B  16K 20. 3  9n
r1  K  A B
1/ 4 1/ 4 n  1/ 2
2. Rate of the reaction decreases with time 21. Both A and R correct only
 from A  products  22. For zero order reaction t completion  a/k
3. Rate does not depends on excess quantity of
23. Rate  K  A 
2
reactant
4. The rate constant becomes double for every
100 C rise in temperature. For 20 C rise, the 24. r1  K  A 
1
0

r2  K  A 
2
rate constant will be 4 times
 K1  4 K 2 or K 2  0.25K1 r3  K  A 
3

d H2  d  I2  1 d  HI 
5. 
dt

dt

2 dt If  A   1 r3  r2  r1

d  A 1 d  B 25. Rate  K  A 
m

6.  
dt 3 dt
6.25 r  K  2.5A 
m

7. Increase in rate of reaction is maximum for the

reaction having the maximum activation energy 26. According to given data  A  is constant
 Ea
8. Slope  w.r.t  B order is 2
2.303R
9. Rate of reaction w.r.t NH3 =Rate of reaction w.r.t 3
27. th completion = 2t1/ 2
4
O2
10. According to stoichiometric relation 15
th completion = 93.75%  4t1/2
11. 40  16
2
 50 
2
 60 
2
 70 
2
 80
28. In zero order the rate of reaction is independent
rate : 5  2 4 = 80
of the concentration of the reactants.
d  N 2  1 d H2  29. Given is elementary reaction, hence
12.  molecularity is 3.
dt 3 dt
30. CH 3COOC2 H 5  NaOH 
LEVEL-I (H.W)
CH 3COONa  C2 H 5OH
r  k CH 3COOC2 H 5  NaOH  RATE OF REACTION & FACTORS
1. The increase in rate constant of a reaction is
Order = 2 molecularity = 2 more when the temperature increases from
31. Molecularity never exceeds ‘3’ 1) 290 K - 300 K 2) 300 K - 310 K
32. Molecularity is a whole number and never 3) 310 K - 320 K 4) 320 K - 330 K
exceeds 3’ d [ B]
Order may be zero or fractional or whole number 2. For 3 A  xB, is found to be 2/3rd of
dt
or negative.
d [ A]
33. Three molecules are participating in elementary . Then the value of ‘x’ is
reaction. dt
1) 1.5 2) 3 3) 1/2 4) 2
1 1 For a chemical reaction Y2  2Z  Product,
34. t1/ 2  , t1/ 2  3.
n 1
a a
1
n 1  1 rate controlling step is Y  Z  Q. If then
2
n2 the concentration of Z is doubled, the rate
3 of reaction will
35.  i  t75%  t50%
1) Remain the same 2) Become four times
2
3) Become 1.414 times 4) Become double
 ii  t100%  2t50%
4. Observe the following reaction 2A  B  C
10 The rate of formation of C is 2.2  103 mol
36. t 90%  t 50%
3
L1 min 1 . What is the value of
t 99%  2t 90%
d A
37. t75%  2t50% 
dt
 in mol L
1
min 1  ?
38. The reaction which is having low half life that 1) 2.2  10 3 2) 1.1 10 3
is fastest reaction. 3) 4.4 103 4) 5.5 103
n 1
t1/' 2  a 2  5. Assertion (A) : The rate of reaction can also
39. "    increase w.r.t its product if one of the
t1/ 2  a1 
products act as catalyst
1 Reason (R) : A catalyst lowers the activation
40. t1/2  energy of reactions.
ka n 1 1) Both (A) and (R) are true (R) is the correct
41. Activation energy forward reaction is greater explanation to (A)
than that of backward reaction .Hence the 2) Both (A) and (R) are true but (R) is not the
reaction is endothermic correct explanation to (A)
42. In graph 2, activation energy for X’ is small. so 3) (A) is true but (R) is false
reaction from X  AB takes place fastly.. 4) Both (A) and (R) are false
43. From C  D , activation energy is lesser than 6. In the reaction 2NO  O 2  2NO 2 , if the rate
that of activation energy from A  B of disappearance of O2 is 16gm.min -1 , then
44. Threshold energy = activation energy + normal
the rate of appearance of NO 2 is
energy.
45. Catalyst increases the rate of reaction by 1) 90 gm.min 1 2) 46gm.min 1
decreasing activation energy . 3) 28gm.min 1 4) 32gm.min 1
7. 2A  B  D  E For this reaction proposed 14. Assertion (A) : The order of a reaction may be
defined as the sum of the powers to which the
mechanism A  B  C  D  slow  ,
concentration terms are raised.
A  C  E  fast  . The rate law expression for Reason (R): Order may be zero, positive,
negative and greater than three.
the reaction is
1) Both (A) and (R) are true (R) is the correct
1) r  K  A  B 2) r  K  A B explanation to (A)
2

2) Both (A) and (R) are true but (R) is not the
3) r  K  A 4) r  K  AC correct explanation to (A)
2

3) (A) is true but (R) is false

4) Both (A) and (R) are false
ORDER OF REACTION
15. A reaction which is of first order w.r.t
8. Which of the following represents the
expression for 3/4th life of first order reaction reactant A , has a rate constant is 6 min 1 .If

2.303 2.303 we start with  A  0.5 mol.L1 , when

1) log 3 / 4 2) log 3
K K would  A  reach the value of 0.05 mol.L1.
2.303 K 1) 0.384 min 2) 15 min
3) log 4 4) log 4
K 2.303 3) 20 min 4) 3.84 min
9. For a reaction A+2B  products, when B is 16. For the reaction A  B  products, it is
taken in excess, then the rate law expression found that order of A is 1 and order of B is 1/
can be written as 2.When concentrations of both A & B are
1) Rate=K[A]1[B]0 2) Rate=K[A]1[B]2 increased four times the rate will increase by
3) Rate=K[A][B] 4) Rate=K[A]2[B]1 a factor
10. The unit of rate constant obeying the rate 1) 6 2) 8 3) 4 4) 16
expression r  K  A  B  is
1 2/3
17. A  B  products, the kinetic data of the
reaction is
1) mole -2/3lit 2/3 time-1 2) mole 2/3 lit -2/3 time-1
A  mole lit 1  B  mole lit 1  Rate  mole.lit 1Sec1 
3) mole-2/3lit -2/3 time-1 4) mole 2/3lit 2/3 time -1
11. For a reaction 3A 
1) 0.5 0.5 2  10 4
 Pr oducts . It is found
that the rate of reaction becomes nine times, 2) 0.5 1.0 1.99  104
if concentration of A is increased three times, 3) 1.0 0.5 2.01 10 4
then the order of the reaction is The order of the reaction is
1) 1 2) 2 3) 3 4) 1.414 1) one 2) zero 3) two 4) fractional
12. The rate constant for a first order reaction is 18. Type - I Type - II
60s 1. How much time will it take to reduce I) SO2Cl2  SO2  Cl2 a) pseudo unimolecular
the initial concentration of the reactant to its reaction
1 / 10 th value? II) 2NO2  2NO  O2 b)first order reaction

1) 3837 sec 1 2) 0.03837 sec 1 III) 2 NO  O2  2 NO2 c) second order

reaction
3) 0.03837 min 1 4) 0.3837 sec 1 
IV) CH 3 COOC 2 H 5  H 2 O  H
13. Units of rate constant of first and zero order
reactions in terms of molarity (M) are respec- CH 3COOH  C2 H 5OH d) third order
tively: The correct matching is
1) I-a, II-b, III-c, IV-d 2) I-b, II-a, III-d, IV-c
1) s 1 , Ms 1 2) s 1 , M 3) Ms 1 ,s 1 4) M,s1
3) I-d, II-c, III-b, IV-a 4) I-b, II-c, III-d, IV-a
 MOLECULARITY 27. Sucrose decomposes in acid solution into
19. For an elementary process, glucose and fructose according to the first
1)The order and the molecularity are identical
order rate law, with t1/2  3.00 hours. What
2)The order is greater than the molecularity
3)The order is lesser than the molecularity fraction of sample of sucrose remains after 9
4)The order is always fractional hours?
20. A  B  products is an elementary reaction. 1) 1.250 M 2) 5.00 M 3) 0.125 M 4) 0.250 M
When excess of A is taken in this reaction,then
COLLISION THEORY
the molecularity and order are respectively 28. Consider the energy profile, for the reaction
1) 2 and 2 2) 2 and 1 x  y  R  S .Which of the following
3) 1 and 2 4) 1 and 1 deduction about reaction is not correct?
21.. Assertion (A) : The order of reaction is equal
to molecularity of simple reactions.
Reason (R) : Molecularity of the
reaction can not be fractional. RS
1) Both (A) and (R) are true (R) is the correct
explanation to (A) x y
2) Both (A) and (R) are true but (R) is not the
correct explanation to (A)
3) (A) is true but (R) is false 1) The energy of activation for the backward
4) Both (A) and (R) are false reaction is 80 KJ
2) The forward reaction is Endothermic
HALF LIFE 3) H for the forward reaction is 20 kJ
22. The product of half life  t1/2  and the squaree 4) The energy of activation for F.R is 60 kJ
29. The following figure denotes the energy
of initial concentration of the reactant (a) is
diagram for a reaction
constant. Then the order of reaction is
1) 2 2) 3 3) 0 4) 1 
23. 50% completion of a first order reaction
takes place in 16 minutes, then the fraction
that would react in 32 minutes from the Reactants
Energy

beginning Products
1) 1/2 2) 1/4 3) 1/8 4) 3/4
24. Out of 300g substance [decomposes as per 1st Reaction coordinate
order]. How much (nearly)will remain after
Then the activation energy of the reverse
18 hr?  t1/2  3hr  reaction is
1) 4.6 gm 2) 5.6 gm 3) 9.2 gm 4) 6.4 gm 1) 2x 2) 2y 3) x  y 4) y  x
25. 75% of a first order process is completed in 30. The energies of activation for forward and
30 min .The time required for 93.75% reverse reactions for A 2  B2  2AB are 180
completion of same process(in hr)? kJ/mol and 200 kJ/mol, respectively. The
1) 1 2) 120 3) 2 4) 0.25 presence of a catalyst lowers the activation
26. The half life of a radio active material is one energy of both (forward and reverse) reac-
hour. What would be the time required for tions by 100 kJ/mol. The enthalpy change of
99.9% completion the reacton A 2  B2  2AB in the presence
1) 5 hours 2) 10hours of a catalyst will be (in kJ/mol)
3) 2 hours 4) 20 hours 1) 300 2) 120 3) 280 4) -20
31. Effective collisions are those in which mol- dx
ecules must: 13. K   A for first order  sec 1
dt
1) Have energy equal to or greater than the
threshold energy dx
K for zero order  M ol. lit  1 .sec  1
2) Have proper orientation dt
3) Acquire the energy of activation 14.Both A and R correct only
4) All of these
2.303 a
15. K  log
LEVEL-I (H.W) - KEY t ax
1)1 2) 4 3) 3 4) 3 5) 1 6) 2 7) 2 16. Rate  K  A  B
8) 3 9) 1 10) 1 11) 2 12) 2 13) 1 14) 2
1
15) 1 16) 2 17) 2 18) 4 19) 1 20) 2 21) 2 R 1  K 1    1
2
22) 2 23) 4 24) 1 25) 1 26) 2 27) 3 28) 1
1 
29) 3 30) 4 31) 4 R 2  K  4  1   4   8
2 
LEVEL-I (H.W) - HINTS 17. For zero order reactions rate is independent of
01 Rate constant is inversly proportional to concentrations.
temperature. 18. See examples of different order reactions
19. For elementary process both order and
1 d A 1 d  B
02   molecularity same
3 dt x dt 20. Molecularity is 2 order is 1
03. Rate  K  Y  Z
1/2
21. For simple reactions order of reaction is equal
to molecularity
1 d  A  d C
04.   1
2 dt dt 22. t 1  a n 1
2

05. Catalyst lowers the activation energy of reactions.

23. t 1 is 16 Min.
d O2  1 d  NO 2 
2

06.    32 Min is equal to two half lifes

dt 2 dt
24. In each half life time the amount becomes half .
07. Rate law depends upon slow step
hence18 hours is equal to 6 half lifes
2.303 a 25. 75% completion equal to 2 half lifes hence
08. K  log
t ax t1/2  15 min .
93.75 % completion equal to 4 half lifes
09. Rate does not depend upon the reactant which is
taken as excess quantity 26. t99.9%  10  t50%
10. Units of K  mole1n .lit n 1.sec1 27. 9 hours = 3t1/2
11. rate  K  Re actant 
n
28. For backward reaction
activation energy = 80 - 40 = 40kj
2.303 a
12 t  log 29. Activation energy for backward reaction =x+ y
K ax 30. For exothermic reactions
1 H  EA of B.W.R  EA of F.W.R
If a  1 then  a  x  
10 31. According to collision theory
 4. CHEMICAL KINETICS
6. The rate of the reaction
LEVEL-II (C.W) CH3COOC2H5 +NaOH CH3COONa+C2H5OH
is given as rate =K[CH3 COOC2H5] [NaOH].
RATE OF REACTION & FACTORS If three times water is added to the reaction
1. For the elementary reaction 2A  C the mixture, the rate of the reaction compared
concentration of A after 30 minutes was to the original rate will be
found to be 0.01 mole/lit. If the rate constant 1 1 1
-2
lit mole-1 sec-1 the 1) rd 2) th 3) th 4) 16 times
o f t h e r e a c t i o n i s 2 . 5 x 1 0

3 9 16
rate of the reaction at 30 minutes is 7. The concentration of reaction decreases from
1) 2.5x10-4 mole lit-1 sec-1 0.2M to 0.05M in 5 minutes. The rate of
2) 2.5x10-6 mole lit-1 sec-1
3) 2.5x10-2 mole lit-1 sec-1 reaction in mole.lit 1.sec 1 is
4) 2.5x10-8 mole lit-1 sec-1 1) 8.3x10-4 2) 0.05 3) 0.0005 4) 0.15
8. A  B KA = 10 e 15 -2000/T

2. For 2 NH 3  Au
 N 2  3H 2 rate w.r.t N 2 is C  D KC = 1014 e-1000/T
2  10  3 M min  1 , then rate w.r.t N 2 after 20 Temperature T; K at which (KA=KC)
min will be (in M min-1) 1) 1000K 2) 2000K
1) 2  103 2)  2  10 3 3) 104 4)  2  10 32000 1000
K 4) K 3)
3. The specific rate of a reaction is 1.5x10-4 lit 2.303 2.303
mole-1sec-1. If the reaction is commenced with 9.From the following data for the
0.2 mole/lit of the reactant, the initial rate ofdecomposition of N2O5 at 300C, find out the
the reaction in mole lit-1sec-1 is rate constant(in min–1). Volume of O2 after
1) 1.5  10 4 2) 3  1 0  5 10 min. of the reaction=90ml. Volume of O2
after completion of the reaction=100ml
3) 6  1 0  6 4) 6  1 0  5 1) 2.303 2) 0.2303 3) 0.02303 4) 23.03
4. For the process 2A  products, rate of 10. If doubling the concentration of the reactant
reaction w.r.t A at 10th second is A increases the rate by 4 times and tripling
2
2  10 M s  1 then rates of same process at the concentration of A increases the rate by 9
5th and 15th seconds (order  0) respectively times, the rate is proportional to
are (in M/s) 1) concentration of A
2) square of concentration of A
1) 101 & 4 102 3) under root of conc. of A
2 2
2) 2.7 10 &1.6 10 4) cube of concentration of A

3) 1.6 102 & 2.7 102 11. Consider a system containing NO2 and SO2
in which NO2 is consumed in the following
4) 2 102 & 2 102
two parallel reactions
5. In the process 2 N 2O5( g )  4 NO2( g )  O2( g ) at
2NO2 
K1
N2O4 ; NO2  SO2 
K2
 NO  SO3
t = 10 rate of reaction w.r.t N 2O5 , NO2 & O2 The rate of disappearance of NO2 will be
respectively are equal to
N 2O5 NO2 O2 1) K1[ NO2 ]2  K 2 [ NO2 ]
1) 500mm/min 400mm/min 200mm/min
2) K1[ NO2 ]2  K 2 [ NO2 ][ SO2 ]
2) 1000mm/min 1000mm/min 500mm/min
2
3) 1000mm/min 2000mm/min 4000mm/min 3) 2 K1[ NO2 ]
4) 400mm/min 400mm/min 400mm/min
4) 2
2 K1[ NO2 ]  K 2 [ NO2 ][ SO2 ]
12. Consider the reaction, 17. Idm 3 2M CH3COOH is mixed with 1dm 3 of
2A  B  Products,When concentration of B
alone was doubled, the rate did not change. 3M ethanol to form ester. The decrease in the
When the concentration of A alone was initial rate if each solution is diluted with an
doubled, the rate increased by two times. The equal volume of water would be
unit of rate constant 1) 2 times 2) 4 times
for this reaction is 3) 0.25 times 4) 0.5 times
1) s 1 2) lit.mol1.s1 18. For a reaction, K  2  1013 e 30000/ RT . When log
1 1
3) Unitless 4) mol. lit .s
K  y  axis  is plotted against 1/T  x  axis  ,
13. For a reaction, the rate constant is expressed
as, K  A.e40000/T slope of line will be.......Cal
The energy of the activation is 30000 30000
1) 2)
1) 40000 cal 2) 88000 cal 4.6 46
3) 80000 cal 4) 8000 cal 30000 30000
14. The reaction 3) 4)
2.303 4.6
CH 3COOC2 H 5  NaOH  19. The rate of a reaction doubles when its
CH 3COONa  C2 H 5OH temperature changes from 300K to 310K.
Activation energy of such a reaction will be
is allowed to take place with initial
(JEE MAINS - 2013)
concentrations of 0.2 mole/lit of each reactant.
If the reaction mixture is diluted with water  R  8.314JK1mol1 and log 2  0.3010
so that the initial concentration of each
1) 48.6kJ mol1 2) 58.5kJ mol1
reactant becomes 0.1 mole/lit. The rate of the
reaction will be 3) 60.5kJ mol1 4) 53.6kJ mol1
1) 1/8 th of the original rate 20. Give the following data for the reaction:
2) 1/4 th of the original rate XYZ
3) 1/2 th of the original rate
4) same as the original rate X  Y  Rate 101 ms 1
15. For the decomposition reaction: 1.0 M 1.0 M 0.25
N 2 O 4 g   2NO 2 g  ; the initial pressure of 2.0 M 1.0 M 0.50
1.0 M 2.0 M 0.25
N 2 O 4 falls from 0.46 atm to 0.28 atm in 30 1.0 M 3.0 M 0.25
minute. What is the rate of appearance of Which one is the rate law equation?
1) Rate  K  X Y 2) Rate  K  X  Y
0 1
NO 2 ?
1) 12  102 atm.min 1 2) 1.2  102 atm.min 1 3) Rate  K  X  Y  4) Rate  K  X  Y 
0 2

3) 1.2  10 2 atm.min 1 4) 1.8  10 1 atm.min 1 21. The activation energy for a reaction is
16. The rate for decomposition of NH3 on 9.0kcal/mol. The increase in the rate constant
when its temperature is increased from 298K
platinum surface is zero order. What are the
to 308 K is:
rate of production of N 2 and H 2 in 1) 10 % 2) 100 % 3) 50 % 4) 63 %
mole.lit 1.sec 1 if K  2.5 104 mole.lit 1.sec 1 22. At 300K rate constant for
1) 3.75 104 ,1.25 104 A  products at t = 50 min is 0.02 s 1 , then
rate constant at t = 75 min and
2) 1.25 104 ,3.75 104 310 K will be (in s 1 )
3) 1.25 104 ,3.75 104
0.04  0.02 
4) 1.25 10 ,3.75 10
4 4 1) 2) 0.04  25 3) 0.04 4)  
25  25 
23. The rate expression for the reaction ORDER OF REACTION
A( g )  B( g )  C( g ) is rate  KC A2 CB1/ 2 . What 28. The decomposition of CH3CHO occurs as
changes in the initial cocentations of A and B CH3CHO(g)  CH4(g)+CO(g), the kinetic
will cause the rate of reaction to increase by data provided is
a factor of eight? [ C H

3
CHO] Rate(mol.lit-1.sec-1)
1.75 x 10-3 2.06 x 10-11
1) C A  2; CB  2 2) C A  2; CB  4
3.5 x 10-3 8.25 x 10-11
3) C A 1; CB  4 4) C A  4; CB 1 7.0 x 10-3 3.30 x 10-10
24. For the reaction system: The rate expression thus can be given as
1) K[CH3.CHO] 2) K[CH3.CHO]2
2 NO( g )  O2( g )  2 NO2( g ) volume is suddenly
3) K[CH3.CHO]3 4) K[CH3.CHO]1/2
reduced to half of its value by increasing the 29. Obseve the following data regarding
pressure on it. If the reaction is of first order
2 NH 3  W
 N 2  3H 2
with respect to O2 and second order with
Pressure(in atm) : 5 10 20
respect to NO, the rate of reaction will Half life (min) : 3.6 1.8 0.9
1) diminish to one - eight of its initial value The unit of K is
2) increase to eight times of its initial value 1) min 1 2) atm . m in  1
3) increase to four times of its initial value 3) ( atm . min)  1 4) atm  2 . min  1
4) diminish to one fourth of its initial value
30. For a first order reation t 0.75 is 1386 seconds,
25. The rate constant of a first order reaction at
then the specific rate constant in sec 1 is.
3 1
27 C is 10 min .The ‘temperaturee
0
1) 103 2) 102 3) 109 4) 105
coefficient’ of this reaction is 2.What is the
31. For N 2O5  2 NO2  1/ 2O2 , it is found that
rate constant  in min 1  at 17 0 C for this
d d
reaction? [ N 2O5 ]  K1[ N 2O5 ], [ NO2 ]  K2 [ N2O5 ] ;
dt dt
1) 103 2) 5  104
d
[O2 ]  K3[ N 2O5 ] then
3) 2  103 4) 102 dt
26. A reaction was found to be second order 1) K1  2K2  3K3 2) 2K1  4K2  K3
with respect to the concentration of carbon 3) 2K1  K2  4K3 4) K1  K2  K3
monoxide. If the concentration of carbon
monoxide is doubled, with everything else 32. For a given reaction of first order, it takes 20
kept as same, the rate of reaction will: min for the cocentration to drop from 1.0 M
to 0.6 M. The time required for the
1) Remain unchanged 2) Tripled cocentration to drop from 0.6 M to 0.36 M
3) Increases by a factor four 4) Doubled will be
27. The initial concentration of cane sugar in 1) more than 20 min 2) less than 20 min
presence of acid was reduced from 0.20 to 3) equal to 20 min 4) infinity
33. The half life of a reaction is 46 minutes when
0.10M in 5 hours and to 0.05M in 10 hours,
the initial concentration of the reactant is 0.4
value of K?  in hr 
1
mole/lit and 92 minutes when the initial
concentration is 0.2 mole/lit. The order of the
1) 0.693 2) 1.386 reaction is
3) 0.1386 4) 3.465 1) zero 2) 0.5 3) 2 4) 1
34. The rate constant of a reaction at 40. The rates of a reaction at different times are
temperature 200K is 10 times less than the given below
rate constant at 400K. What is the activation Time (in min) Rate
energy  E a  of the reaction? (R= Gas
0 2.8 x 10-2
10 2.8 x 10-2
constant) 20 2.8 x 10-2
1) 1842.4R 2) 921.2 R 30 2.79 x 10-2
3) 460.6 R 4) 230. 3 R The order of the reaction is
35. The decomposition of ozone proceeds as 1) 2nd order 2) zero order
O3  O2  O (fast) 3) 3rd order 4) 1st order
41. The isomerization of cyclopropane to form
O  O3  2O2 (slow)
propene(   CH 3  CH  CH 2 ) is a first
the rate expression should be
order reaction. At 760K, 85% of a sample of
1) Rate  K O3  2) Rate  K O3  O2 
2 2 1
cyclopropane changes to propene in 79 min.
Calculate the value of the rate constant.
3) Rate  K O3 O2  4) Rate  K O3 O2 
1
1) 2.42 min 1 2) 3.66 102 min1
36. 50% of a reaction is completed in 16min, 3) 2.40 102 min1 4) 1.04 102 min1
under similar conditions 75% of the reaction 42. For a reaction following first-order kinetics,
is completed in 48min. Order of the reaction which of the following statements are
will be correct?
1) 3 2) 1 3) 2 4) 0 1) The time taken for the completion of 75% of
37. For a process, A  B  products, the rate of
the reaction is twice t1/2.
the reaction is second order with respect to A
and zero order with respect to B. When 1.0 2) A plot of the reciprocal of the concentration
mole each of A and B are taken in a one litre of the reactants against time gives a straight line
3) The degree of dissociation is equal to 1ekt
vessel, the initial rate is 1102 mol. lit 1. s1 .
4) A plot of  A0 /  A versus time gives a straight
The rate of the reaction, when 50% of the
reactants have been converted to products line.
would be 43. t 1 / 4 can be taken as the time taken for the
1) 1102 mol.lit 1 . s1 concentration of a reactant to drop to 3/4 of
its initial value. If the rate constant for a first
2) 2.5103 mol. lit 1. s1
order reaction is K, t1/4 can be written as
3) 5.0 102 mol.lit 1.s1 1) 0.75/K 2) 0.69/K 3) 0.29/K 4) 0.10/K
4) 0.5102 mol. lit 1.s1 44. The rate constant is numerically the same for
three reactions of first, second and third
38. SO2 Cl2  SO2  Cl2 is a first order gas order respectively. Which of the following is
reaction with K  2.2 105 sec1 at 3200 C . correct:
The percentage of SO2Cl2 decomposed on 1) if  A   1 then r1  r2  r3
heating for 90 minutes is: 2) if  A   1 then r1  r2  r3
1) 1.118 2) 0.1118 3) 18.11 4) 11.18
39. For the reaction a A  x P when [A] = 2.2 mM 3) if  A   1 then r3  r2  r1 4) All of these
the rate was found to be 2.4 mM s 1 . On 45. 99% of a 1st order reaction completed in
reducing concentration of A to half, the rate 2.303 minutes. What is the rate constant and
1
changes to 0.6 m M s . The order of reaction half-life of the reaction
with respect to A is 1) 2.303 and 0.3010 2) 2 and 0.3465
1) 1.5 2) 2.0 3) 2.5 4) 3.0 3) 2 and 0.693 4) 0.3010 and 0.693
46. For a first order reation, (A)  products, the 53. An endothermic reaction A  B has an
concentration of A changes from 0.1 M to activation energy 15 K Cal/mole and enthalpy
0.025 M is 40 min. The rate of reaction when change ( H ) of the reaction is 5 KCal/mole.
the concentration of A is 0.01 M is The activation energy of the reaction B  A
( AIE - 2012 ) is
 
1) 3.47  10 M min 2) 3.47  10 M min
4 1  5  1 1) 20 K cal/mole 2) 15 K cal/mole
3) 10 K cal/mole 4) 5 Kcal/mole
3) 1.73  104 M min 1 4) 1.73  105 M min 1
47. The reaction 2A  B is first order in A with LEVEL-II (C.W) - KEY
a rate constant of 2.8  10 2 s 1. How long will 1) 2 2) 1 3) 3 4) 2 5) 4 6) 3 7) 3
it take(nearly) for A to decrease from 8) 4 9) 2 10) 2 11) 4 12) 1 13) 3 14) 2
0.88 M to 0.22M ? 15) 3 16) 2 17) 3 18) 4 19) 4 20) 3 21) 4
1) 60 s 2) 76 s 3) 50 s 4) 44 s 22) 3 23) 2 24) 2 25) 2 26) 3 27) 3 28) 2
HALF - LIFE 29) 2 30) 1 31) 3 32) 3 33) 3 34) 2 35) 2
48. For a first order reaction with half-life of 150 36) 3 37) 2 38) 4 39) 2 40) 2 41) 3 42) 1
seconds, the time taken for the concentration 43) 3 44) 4 45) 2 46) 1 47) 3 48) 3 49) 2
of the reactant to fall from M / 10 to M / 100 50) 3 51) 2 52) 2 53) 3
will be approximately
1) 1500 sec 2) 900 sec LEVEL-II (C.W) - HINTS
3) 500 sec 4) 600 sec
49. In a first order reaction, 50 minutes time is 1. r  K  A
2

taken for the completion of 93.75% of a 2. Given reaction is Zero order reaction.
reaction. Half life of the reaction is For zero order rate remains constant with time.
1) 25 min 2) 12.5 min 3) 20 min 4)10 min 3. From units of rate constant, it is second
50. In 69.3 min, a first order reaction is 50% order Rate  K  reactant 
2

complete. How much reactants are left after

161 min? 4. Rate of reaction decreases with time
1) 80% 2) 40% 3) 20% 4) 60% 5. Rate of reaction is same for all the reactents and
51. Type - I Type - II products
1 6. r  K CH 3COOC2 H 5  NaOH  on adding,
I) first order reaction a) 
an1 three times of water the resultant volume
II) zero order reaction b) Radio active decay becomes four times that of initial volume.
III)trimolecular reaction c) Photochemical change in conc.
reactions 7. Rate= time interval
IV)half life period of d) 2CO  O 2  2CO 2
‘n’ th order 1015. e2000/T  1014 e1000/T
8.
1) I - a, II - b, III - c, IV - d 10  e1000/T
2) I - b, II - c, III - d, IV - a
3) I - c, II - d, III - b, IV - a 2.303 a
9. K  log
4) I - d, II - c, III - b, IV - a t ax
COLLISION THEORY 10. 2n  4, 3n  9
52. For A+B  C+D H =-20 kJ mole -1. The rate  K1  NO2      1
2

activation energy for the forward reaction is 11.

85 kJ. Then the activation energy for the rate  K 2  NO2 SO2       2 
1 d  NO 2  d  NO 2 
backward reaction is ___
  K1  NO 2  ;  K 2  NO 2 SO 2 
2
1) 65 kJ 2) 105 kJ 3) 85 kJ 4) 40 kJ 2 dt dt
12. r  K  A1  B0 22. For every 100 C temperature rate constant
First order : unit of K = sec1 almost doubled
23. Rate  K A B
2 1/2
13. K  Ae Ea / RT
14. r  K CH 3COOC2 H 5  NaOH  24. Rate  K  NO  O2 
2

1 d  NO 2   0.28  0.46  25. For every 100 C temperature rate constant

15. 
2 dt 30 almost doubled
1/ 2d  NO 2  26. Rate  K  CO
2
 6  103
dt 27. Given reaction is first order reaction
d  NO 2  0.693
 1.2  102 atm.min 1  t1/2 
dt K
1 d  NH 3  d  N 2  1 d  H 2  28. From result (1) & (2) , as the concentration
16.    increases by 2 times , rate becomes 4 times. so
2 dt dt 3 dt
it is second order.
Rate  K  NH3 
0
29. Given reaction is zero order reaction

d  NH 3  30. t75%  2t50%

 2.5  104 mol.ltr 1S1
dt 0.693
t50% 
d  N2  1 K
  2.5  104  1.25  104
dt 2 d d d O 
31. 2   N 2O5    NO2   4 2
d H2  3 dt dt dt
  2.5  104  3.75  104 32. For 0.4M concentration decreasing 20 min time
dt 2
required. for 0.24 M concentration decreasing
CH3COOH  C2H5OH  CH3COOC2H5  H2O lessthan 20 min required
17.
 2 moles 3moles t '1 n 1
a 
rate  K CH3COOHC2 H5OH  2  2
33. From given data t  a1 
''

18. y  mx  c 1
2

30000 30000
Slope  m   so the reaction is the second order
2.303 R 2.303 2
K E 1 1
34. log K  2.303R  T  T 
2 a
K 2 Ea  T2  T1   1 2
19. 2.303log   
1

K1 R  T1T2 
35. O  O3 
slow
 2O2
20. rate  K  X Y 
0
r  K1 O O3 
K c O3 O3 
K 2 Ea  T2  T1   K1
21. 2.303log    O2 
K1 R  T1T2 
 K O3  O2 
1
O3  O2  O
2
K 9  10 
 2.303log 2 
K1 2 103  298  308  O2 O 
Kc 
K2 O3 
  1.63 ; i.e. 63% increase
K1 36. For second order : t 2/3  2t1/2
37. K  1102 lit. mol 1.s1 2.303 a
45. K  log
rate = K[ A]2[B]0 t ax
When 50% of the reactants are converted into 46. Rate = KA  0.003477  0.01
products
2.303 a
rate = 1102 (0.5)2  2.5 103 mol .lit 1.s1 47. K  log
t ax
2.303 a
38. Use K  log to then with the help 48. t  0.693 , K  2.303 log a
t ax 1/2
K t ax
of this calculate x
49. 93.75% completion 4t1/ 2 are required
39. r  K  A n ......(1)
50
r A
n
Total time required is 50 min ;  t1/2 
 K   ......(2) 4
4 2
50. t1/2  69.3min for 2t1/2  138.6 min
 2
 4  2n For 161 min 20% is left
1 51. All radioative disintegration reactions are first
n=2 1
40. For zero order reactions, the rate of the reacion order and t1/2 
a n 1
does not change with time.
52. Activation energy for backward reaction
2.303 a = H  activation energy for forward reaction.
41. K  log
t ax 53. Activation energy for backward reaction
2.303 100 = H  activation energy for forward reaction.
 log
79 100  85
K  2.4 102 min 1
LEVEL-II (H.W)
42. For first order t75%  2  t50%
RATE OF REACTION & FACTORS
2.303 a
43. K  log 1. For a reaction Ea  0 and K  3.2  104 s 1 at
t ax
300 K. The value of K at 310 K would be
2.303 a 2.303 4 1) 6.4  10 4 s 1 2) 3.2  10 4 s 1
 log  log
t1/2 3a / 4 t1/4 3
3) 3.2  108 s 1 4) 3.2  105 s 1
2.303  0.125 0.29 2. The rate constant, k for the reaction
K 
t1/4 t1/4 1
N 2 O5  g   2NO 2  g   O 2  g 
44. 2
is 2.3  10 2 s 1 . Which equation given below
r1  K  A 
1
describes the change of  N 2 O5  with time ?
r2  K  A 
2
 N2 O5 0 and  N2 O5 t correspond to
r3  K  A 
3
concentration of N 2 O5 initially and at time t
1)  N 2 O5 t   N 2 O5 0  kt 2)  N 2 O5 0   N 2 O5 t ekt
If  A   1 r3  r2  r1
3) log10  N 2 O5 t  log10  N 2 O5 0  kt
If  A   1 r1  r2  r3
N O 
4) ln  N O   kt
2 5 0

If  A   1 r1  r2  r3 2 5 t
3. The rate reaction for the reaction 2A  B  C 9. For 2A  B  C  products, rate law is given
is found to be rate =K[A] [B]. The correct
by rate  K  A  B & rate constant (K) is
2
statement is relation to this reaction is that
the 2 106 M 2 . S 1 . Then rate of the reaction
1
1) unit of K must be s
become 2 10 9 M 2 . S 1 only when
2) value of K is independent of the initial
concentration of A and B A  B  C
3) rate of formation of C is twice the rate of
1) 0.3 M 0.2 M 0.2 M
disappearance of A.
2) 0.2 M 0.1 M 0.2 M
4) t1/ 2 is a constant 3) 0.1 M 0.1 M 0.1 M
4. The rate law for a reaction between the 4) 0.2 M 0.2 M 0.1 M
substances A and B is given by Rate= K An Bm
ORDER OF REACTION
on doubling the concentration of A and halving 10. Rate expression for xA + y B
 products is
the concentration of B, the ratio of the new
rate to the earlier rate of the reaction will be Rate  K[ A]m [ B]n . Units of K w.r.t A and B
as respectively are s 1 and M 1.s 1 when
1 concentrations of A and B are increased by 4
1)  m n  2) (m + n) 3) (n – m) 4) 2 n  m  times, then
2
5. Hydrogenation of vegetable ghee at 250 C 1) R f  16 Ri 2) Ri  16 R f
reduces pressure of H 2 from 2 atmospheree 3) R f  8 Ri 4) R f  64 Ri
to 1.2 atmosphere in 50 minutes. The rate of 11. A first order reaction was commenced with
reaction in terms of molarity per second is 0.2 M solution of the reactants. If the molarity
1) 1.09  106 2) 1.09  105 of the solution falls to 0.02M after 100
3) 1.09  10  7 4) 1.09  10  8 minutes the rate constant of the reaction is
1) 2 x 10-2 min-1 2) 2.3 x 10-2 min-1
6. For a reaction 2 SO2  O2  2 SO3 rate of
3) 4.6 x 10-2 min-1 4) 2.3 x 10-1 min-1

consumption of SO2 is 6.4  10 kg / sec. the 12. The experimental data for the reaction
3

rate of formation of SO3 in same units will 2NO g   Cl2 g   2 NOCl g  are given below
be
Expt Cl2   NO  Initial rate
1) 6.4 103 2) 8 103 3) 4 103 4) 16 103
7. A gaseous reaction 1 0.020 0.010 2.4  10 4
1 2 0.020 0.030 2.16  10 3
A2 ( g )  B ( g )  C (g) shows increase of
2 3 0.040 0.030 4.32  10 3
pressure from 100mm to 120mm in 5 min. What is the order of the reaction ?
The rate of disappearance of A2 is 1) 1 2) 2 3) 3 4) 0
1) 4mm min-1 2) 40mm min-1 13. The reaction, 2A  B  Products, follows the
3) 8mm min-1 4) 20mm min-1 mechanism
8. Concentration of a reactant ‘A’ is changed
2 A  A2
from 0.044M to 0.032M in 25 minutes, the
average rate of the reaction during this A2  B  Products (Slow) The order of the
interval is reaction is
1) 0.0048 mol.lit 1.min 1 2) 0.00048 mol.lit 1.sec1 1
1) 2 2) 1 3) 3 4) 1
3) 4.8 104 mol.lit 1.min 1 4) 0.0048 mol.lit 1.sec 1 2
14. Diazonium salt decomposes as HALF LIFE
C6 H 5 N 2 Cl  C6 H 5 Cl + N 2 18. The half-life for the reaction.
1
At 0o C , the evolution of nitrogen becomes N 2 O5  2NO 2  O 2 is 24hrs. at 300 C.
2
two times faster when the initial
concentration of the salt is doubled therefore Starting with 10g of N 2 O5 how much grams
it is. of N 2 O5 will remain after a period of 96
1) a first order reaction. hours?
2) a second order reaction. 1) 1.25 g 2) 0.625 g 3) 1.77 g 4) 0.5 g
3) independent of the initial concentration of the 19. The half life of a first order reaction
salt. A  B  C is 10 minutes. The concentration
4) a zero order reaction of ‘A’ would be reduced to 10% of the original
15. Using the data given below the order and rate concentration in
constant for the reaction : 1) 10 minutes 2) 33 minutes
CH 3CHO  g   CH 4  g   CO  g  would be 3) 90 minutes 4) 70 minutes
Experiment Initial conc. Initial rate 20. A first order reaction is half-completed in 45
No (mol/l) (mol/l) minutes.How long does it need for 99.9% of
a 0.10 0.020 the reaction to be completed?
b 0.20 0.080 1
1) 7 hours 2) 20 hours
c 0.30 0.180 2
3) 10 hours 4) 5 hours
d 0.40 0.320
Answer is COLLISION THEORY
21. The reaction A  C has activation energy for
1) 2,  K  2.0 l / mol sec 2) 0,  K  2.0 mol / l sec
the forward and the backward reaction has
25KJ and 32KJ respectively. The  H for
3) 2,  K  1.5 l / mol sec 4) 1,  K  1.5 sec 1 
the reaction is
16. For the first order reaction A  product. 1) 57 KJ 2) -57 KJ 3) 7 KJ4) -7 KJ
22. Consider an endothermic reaction X  Y
When the concentration of A is 2.5  10 2 M
with the activation energies E b and E f for the
the activation energy is 20K.Cal/mole. If the
backward and forward reactions
conc. of  A  is doubled, at same temperature, respectively. In general.
the activation energy becomes equal to 1) E b  E f 2) E b  E f 3) E b  E f
4) There is no definite relation because E b and E f
1) 40K.cal/mole 2) 10K.cal/mole

3) 20K.cal/mole
20
4) 2 RT K.cal/mole LEVEL-II (H.W) - KEY
1) 2 2) 4 3) 2 4) 4 5) 2 6) 2 7) 3
17. A g   B g  is a first order reaction. The initial 8) 3 9) 3 10) 4 11) 2 12) 3 13) 3 14) 1
concentration of A is 0.2 mol.lit 1. After 10 15) 1 16) 3 17) 1 18) 2 19) 2 20) 1 21) 4
minutes the concentration of B is found to be 22) 1

0.18 mol.lit 1. The rate constant  in min  for

1
LEVEL-II (H.W) - KEY
the reaction is 1. If E a  0 then k = A = constant
2. Decomposition of N 2 O5 is of first order
1) 0.2303 2) 2.303 3) 0.693 4) 0.01
3. Given reaction is second order
4. r  K  A   B    (1)
n m 16. It is first order reaction
2.303 a
1 
m
17. K  log
r1  K  2A   B    (2) ax
n
t
2 
18. 24 hours = 1t1/2
 2  r1 n m
 2 ; r1  2n  m.r
1 r 96 hours = 4t1/ 2
5. P  0.8atm / 50 min 19. For 3t1/2 concentration decreases to 12.5%
n P 20. t 99.9%  10t 50%
PV  nRT ; P  V RT ; M  RT
21. For exothermic reactions
6. According to Stoichiometric relation H  EA of B.W.R  EA of F.W.R
1
7. A2  B  C 22. For endothermic reaction activation energy of
2
forward reaction is greater than activation energy
x of backward reaction.
100  x x
2
x PREVIOUS EAMCET QUESTIONS
100   120 1. In a first order reaction the concentration of
2
x  40 the reactant decrease from 0.6 M to 0.3 M in
15 minutes. The time taken for the
dx 40 concentration to change 0.1 M to 0.025 M in
Rate =   8mm.min 1
dt 5 minutes is. (2014-E)
1) 1.2 2) 12 3) 30 4) 3
 0.044  0.032 
8. Average rate    2. Which of the following plots is correct for a
 25 first order reaction (2013-E)
9. According to given data C is independent
10. Acc to units w.r.t A is first order w.r.t B is second
order
Log(a-x)
2.303 a
11. K 
Log(a-x)
log
t ax 1) 2)
12. W.r.t NO order is 2 and w.r.t Cl2 order is 1 Time Time

13. r  K [ A2 ][ B] .............. (i)

[ A2 ]
K (a-x)
[ A]2 Log(a-x)

3) 4)
 [ A2 ]  K [ A] .................. (ii)
2
Time Time
From (i) and (ii)
r  KK[ A]2 [ B]  K '[ A]2 [ B] 3. Which one of the following statements is
14. r  K  A ; 2 r  K  2 A  n 1 correct for the reaction (E-2012)
n n

CH 3COOC 2 H 5  NaOH 
 CH 3COONa  C 2 H 5OH
15. r  K CH 3CHO   (1)
n (aq) (aq)
 aq   aq 

1) Order is two but molecularity is one

4r  K  2CH 3CHO   (2)
n
2) Order is one but molecularity is two
1 x
3) Order is one but molecularity is one
(2)
n2 K 4) Order is two but molecularity is two
(1) at (a  x)
4. Match the following : (2011-E) 9. For the reaction A  3B  2C  D, which one
LIST-I LIST - II of the following is not correct? (M-2010)
A)Arrhenius equation i)free energy change 1) Rate of disappearance of A= Rate of
B)Slowest step in a ii) conc-1 .time-1 formation of D
reaction mechanism
2
C)Rate constant of a II order iii) conc t  n .time 1 2) Rate of formation of C   Rate of
D)The possibility of a iv) Rate determining 3
reaction step disappearance of B
1
v) k = Ae. Ea /RT 3) Rate of formation of D   Rate of
1) A-v, B-i, C-iii, D-iv 2) A-v, B-iv, C-iii, D-ii 3
3) A-v, B-iv, C-ii, D-i 4) A-iii, B-iv, C-ii, D-i disappearance of B
5. The plot of log K vs 1/T yields a straight line. 4) Rate of disappearance of A  2  Rate of
The slope of the line would be equal to formation of C
(E-2010)
Ea Ea Ea Ea
PREVIOUS EAMCET - KEY
1)  2)  3) 4) 1) 3 2) 3 3) 4 4) 3 5) 2 6) 3
R 2.303R R 2.303R
6. For a first order reaction at 27 0 C the ratio 7) 2 8) 3 9) 4
of time required for 75% completion to 25%
completion of reaction is (2009-E)
HINTS
1) 3.0 2) 2.303 3) 4.8 4) 0.477 t 1  15
1.
7. The rate constant of a first order reaction is 2

0.0693min 1 .What is the time required (in 0.1 

15
 0.05 15
 0.025
minutes) for reducing an initial concentration Two half life = 30 min
of 20mol lit -1 to 2.5mol lit -1 (2009-M) 1
2. log  a  x  is directly proportionl to for first
1) 40 2) 30 3) 20 4) 10 T
8. For a reversible reaction A  B which one order
of the following statements is wrong from the 3. Saponification of ester is a second order reaction
given energy profile diagram. (2008-E) and molecularity of the reaction is two.
Rate  K  ester   NaOH 
1 1

4. According to arrhenius equation and reaction

E mechanism.
Ea
A
B 5. log K  log A 
2.303RT
y  c  mx
2.303 a
Reaction coordinate 6. K log
t ax
1) Activation energy of forward reaction is 2.303
greater than backward reaction. For 75% completion  t1 log 4
2) The forward reaction is endothermic K
3) The threshold energy is less than that of 2.303 4
For 25% completion  t 2  log
activation energy K 3
4) The energy of activation of forward reaction t1 log 4
is equal to the sum of heat of reaction and the   4.8
t 2 log 4
energy of activation of backward reaction. 3
 0.693 0.693 d  A
7. t1/2    10 min 4. For the reaction A  Products,  k
K 0.0693 dt
20  t1/2
10 t1/2
5 
t1/2
 2.5 and at different time interval, [A] values are
3  t1/2  3 10  30min Time 0 5 min 10 min 15 min
8. An inspection of the energy profile diagram [A] 20 mol 18 mol 16 mol 14 mol
indicates that 1,2 and 4 are correct statements 3 At 20 minute, rate will be:
is wrong statement. 1) 12 mol/min 2) 10 mol/min
d  A  1 d  B 1 d  B d  D  3) 8 mol/min 4) 0.4 mol/min
9.    5. The decomposition of 2N2O5 2N2O4 O2 is at
dt 3 dt 2 dt dt
200C . If the initial pressure is 114 mm and
LEVEL-III after 25 min. of the reaction the total pressure
of gaseous mixture is 133 mm. Calculate the
RATE OF REACTION AND FACTORS average rate of the reaction in

d  A d  B d C 
 a  Atm.m1  b  mol s1 respectively
1. xA  yB  zC. If    1.5 1) 0.002, 8.58 107 2) 0.001,8.58  10 7
dt dt dt
then x,y and z are: 3) 0.002,8.58  104 4) 0.001,8.58  103
1) 1,1,1 2) 3,2,3 3) 3,3,2 4) 2,2,3 6. A  B & C  D are first order reactions,
2. For the complex
ratio of t99.9% values is 4 : 1, then ratio of rate

Ag  2 NH 3   Ag  NH 3 2 

constans K1 to K2 is
1) 4 : 1 2) 2 : 1 3) 1 : 1 4) 1 : 4
 dx 
 dt 

   2  10 L mol s  Ag   NH 3 
7 2 2 1  2
 7. A substance ‘A’ decomposes in solution
following first order kinetics. Flask I contains
 1L of a 1M solution of A and flask II contains
1 102 s 1  Ag  NH 3 2  100 ml of a 0.6 M solution. After 8 hours the
Hence, ratio of rate constants of the forward concentration of A in flask I has become 0.25.
What will be the time taken for concentration
and backward reaction is:
of A in flask II to become 0.3 M?
1) 2  107 L2 mol 2 2) 2  109 L2 mol 2 1) 0.4 h 2) 2.4 h 3) 4.0 h
3) 1102 L2 mol 2 4) 0.5 109 L2 mol 2 4) Can’t be caculated since rate constant is not
3. The following reaction is first order in A and given
first order in B: 8. The energy of activation for a reaction is
50kJ/mol. Presence of a catalyst lowers the
A  B  Product, Rate  kI A B  energy of activation by 25%. What will be
the effect on rate of reaction at 300C . Other
things remains same.
A 1) 142.75 2) 242.75 3) 342.75 4) 442.75
9. For the reaction 2 A  3B  product, A is in
B
excess and on changing the concentration of
B from 0.1 M to 0.4 M, rate becomes doubled.
I II Thus, rate law is:
dx dx
 k  A  B   k  A B 
2 3
Relative rate of this reaction in vessel I and 1) 2)
II of equal volume is: dt dt
A dx dx 1
1) 1:1 2) 1:2 3) 2:1 4) 1:4  k  A  B   k  B 2
0 2
3) 4)
B dt dt

II
10. A reaction is catalysed by H+ ion; in 15. For the reactions of I, II and III
3 1
presence of HA rate constant is 2  10 min orders, K1  K 2  K 3 when concentrations aree
and in presence of HB rate constant is expressed in mol litre1 . What will be
3 1
1 10 min . HA and HB (both strong acids) the relation in K1 , K 2 , K 3 ,if concentrations aree
have relative strength as: expressed in mol/mL ?
1) 0.5 2) 0.002 3) 0.001 4) 2
1) K1  K2  K3 2) K1  K2 103  K3 106
ORDER AND HALF LIFE
3) K1  2K2  K3 4) 2K1  3K2  4K3
16. 900 ml of pure and dry O2 is subjected to
seilent electric discharge, so that after a time
10 min. volume of ozonized oxygen is found
to be 870 ml. Now average rate of reaction in
this interval is (in ml/min)
1) 3 2) 9 3) 90 4) 60
11.
17. At some temperature, the rate constant for
the decomposition of HI on the surface of gold
is 0.08 MS1 2HI  g   H 2  g   I2  g  what is the
Half-life is independent on concentration of order of the reaction. Howlong will it take
A. After 10 minutes volume of N2 gas is 10 L for the concentration of HI to drop from 1.50
and after completion of reaction 50 L. Hence, M to 0.30 M.
rate constant is : 1) zero order, t  7.5Sec
2.303 2.303 2) zero order, t=15 Sec
1) log 5 min 1 2) log1.25 min 1
10 10 3) first order, t=22.5 Sec
2.303 2.303 4) first order, t=7.5 Sec
3) log 2 min 1 4) log 4 min 1 18. For the first order gaseous reaction:
10 10
12. When the initial concentration is changed x ( g )  2 y ( g )  z ( g ) the initial pressure,

from 0.50 to 1.0 mole lit , the time of half Px  90 mm Hg . The pressure after 10
1

completion for a certain reaction is found to

minutes is 180 mm Hg. The rate constant of
change from 50 sec. to 25 sec. Calculate the
the reaction is:
order of reaction.
1) 1 2) 2 3) 3 4) 0 1) 2  10 3 sec 1 2) 2  103 sec 1
13. For SO2Cl2( g )  SO2( g )  Cl2( g ) pressures of 3) 1.15  103 sec 1 4) 1.15  103 sec 1
SO2Cl2 at t = 0 and t = 20 minutes respectively HALF LIFE
are 700 mm and 350 mm. When log ( P0 / P )
is plotted against time (t), slope equals to 19. For the zeroth order reaction, sets I and II
are given, hence x is :
1) 1.505  10 2 s 1 2) 1.202  10 3 min 1
3) 1.505  10 2 min 1 4) 0.3465 min 1 I:
dX n
t=0 t = 2 min
14. For a reaction  K  H   . If pH of reaction
dt
medium changes from two to one, the rate II :
becomes 100 times of the value at pH = 2. t=0 t = x min
The order of reaction is
1) 1 2) 2 3) 0 4) 3 1) 2 min 2) 3 min 3) 4 min 4) 6 min
20. A  B , K1  0.693sec 1 26. The radioactive isotope 32 P decays by first
order kinetics and has a half-life of 14.3 days.
C  D, K 2  0.693min 1 . If t1 & t2 are half
How long does it take for 95.0% of a given
lives of two reactions, then
sample of 32 P to decay?
1) t1  t2 2) t1  60t2 1) 21 days 2) 42 days 3) 62 days 4) 80 days
3) t2  60 t1 4) t2  2.303t1 27. 2 0 % d e c o m p

2
O2 in presence of
o s i t i o n o f H

an acid requires 5 min. The time required for

21. For 2 NH 3( g )  Pt

 products follows zeroo 50% decomposition in minutes is
order kinetics. If t1/ 2 at p = 4 atm is 25 sec, 1) 15.52 2) 1.552 3) 0.1552 4) 7.76
28. The first order rate constant for the
t1/ 2 at p = 16 atm will be (in sec)
decomposition of N 2 O5 is 6.2 104 s 1 . The
1/ 4
1) 6.25 2) 625 3) 100 4) (25)
half-life period for this decomposition is :
22. DDT on exposure to water decomposes
1) 223.4 s 2) 1177.7s 3) 1117.7 s 4) 160.9 s
according to first order kinetics. Half life =
10 years. How much time it will take for its ARHENIUS EQUATION &
decomposition to 99%? COLLISION THEORY
1) 50 years 2) 66.6 years 29. The rate constant of a reaction at 300 K is
3) 500 years 4) 666 years
1.6x10-3 sec-1 and at 310K it is 3.2x10-3 sec-1
23. The rate of a first order reaction is 0.04 mol
the activation energy of the reaction
litre–1 sec–1 at 10 minute and 0.03 mol litre–1
approximately in kcals is
sec–1 at 20 minute after initiation. Find the
1) 12-13 2) 20-25 3) 30-40 4) 40-50
half life of the reaction.
30. The rate constant, the activation energy and
1) 2.406 min 2) 24.06 min
3) 240.6 min 4) 0.204 min the Arrhenius parameter of a chemical
24. A  B  product. reaction at 25C are 3  104 s 1 , 104.4 KJ / mol
from the following data calculate half life and 6 1014 s1 respectively. The value of the
period. rate constant as T   is
 A0   B0  Rate  R 0  1) 2.0 1018 s 1 2) 6.0 1014 s 1
mol litre–1 mol litre–1 mol litre–1 sec–1 3) Infinity 4) 3.6 1030 s 1
0.1 0.2 0.05 31. For an exothermic chemical process occuring
0.2 0.2 0.10 in two steps as
0.1 0.1 0.05
 i  A  B  X  slow   ii  X  AB  fast 
1) 1.386 sec 1 2) 13.86 sec 1
The progress of the reaction can be best
3) 26.72 sec 1 4) 2.672 sec 1 described by
25. The gas phase decomposition of dimethyl
ether follows first order kinetics:
CH3  O  CH3 ( g)  CH4 ( g)  H2 ( g)  CO( g)
1) 2)
The reaction is carried out in a constant
volume container at 500 0 C and has a half
life is 14.5 minutes. Initially only dimethyl
ether is present at a pressure of 0.40 atm.
What is the total pressure of the system after 3) 4) None
12 minutes? (Assume the ideal gas
behaviour.)
1) 0.946 atm 2) 0.785 atm
3) 0.777 atm 4) 0.749 atm
32. Given that the temperature coefficient for the 38
saponification of ethyl acetate by NaOH is Average rate =  0.002 Atm m 1
25  760
1.75. Calculate the activation energy.
Again PV = nRT
1) 1.0207 kcal 2) 10.207 kcal/mol.
3) 1.0207 cal 4) 10.207 cal/mol. n 0.002
 
33. What is the activation energy for the V 0.0821 473  60
decomposition of N 2O5 as, t1 K2
6. 
1 t2 K1 for first order reaction
N 2O5  2 NO2  O2 if the values of the rate
2 7. The concentration of A remains 1/4th in 8 hours.
Therefore 1/  2 , n  2
n
constants are 3.45  105 and 6.9 103 at
27 0 C and 67 0 C respectively? and E1/ 2 is 8  n  t1/ 2 ; t1/ 2  4; In 4 hours 0.6
1) 112.5 kJ 2) 200 kJ 3) 149.5 kJ 4) 11.25 kJ will become 0.3
8. K  Ae Ea / RT
LEVEL-III - KEY
K1  Ae5010 / RT
3

1) 3 2) 2 3) 2 4) 4 5) 1 6) 4 7) 3
K2  Ae37.510 / RT
3
8) 1 9) 4 10) 4 11) 2 12) 2 13) 3 14) 2
15) 2 16) 2 17) 2 18) 3 19) 4 20)3 21)3
K1
e 
 37.5 50 103 / RT
 e12.510 / RT
3

22) 2 23) 2 24) 1 25) 4 26) 3 27) 1 28) 3 K2

29) 1 30) 2 31) 3 32) 2 33) 1
K 2 12.5 103
2.303log 
LEVEL-III - HINTS K1 8.314  303
1 d  A  1d  B 1 d  C K2 r2 K2
 
1. 3 dt 3dt 2 dt  142.75   142.75
K1 r1 K1
Kf 2  107 9. Rate of reaction does not depends on excess
2. 
K b 1 10 2 quantity
3. When compared to vessell-1, In vessel-II the 2
concentration of A doubled. therefore rate 10. Relative strength = ratio of rate constants = 1
becomes two times. 11. T50 is independent of concentration of A. Hence
4. For every 5 minutes change in concentration is first-order reaction.
2 moles per litre
A  N2  g 
2
Rate   0.4 At t = 0, a 0
5
At time t,(a-x) x = 10L
5. 2N 2 O5  2N 2 O 4  O 2 (after 10 minutes)
a x x x /2
At complete
Pressure due to a moles = 114 mm Reaction, (a-a) a = 50 L
Pressure due to a 
x
moles = 133 mm   a  x   40 L
2
2.303 50
x k  log
Hence pressure due to moles = 133 - 114 10 40
2
2.303
and due to x moles 19  2  38 mm  log1.25 min 1
10
 Pressure due to (a – x) moles = 114–76 = 38.
 T1  a 2 
n 1 Given, K1  K 2  K3
12.  
T2  a1  K1' K '2 K3'
  
50  1.0 
n 1 103 106 109
2   2 or 21   2 
n 1 n 1
 
25  0.5  or K1'  K '2 103  K 3' 106
 1  n  1 . Hence n = 1 + 1 = 2. Reaction is of 16. 3O2  2O3
second order.
a  3x 2x
2.303 P
13. K  log 0 t  10 min 900  3x 2x
t P total volume after 10 min  900  x
2.303 700 900  x  870
K log  0.03466
20 350 x  30 ml
P0 Kt  volume O2 consumed in 10 min is 3x mean
log  0
P 2.303 90 ml
y  mx  c
dc 90
0.03466    9 ml / min
 m  0.01505 dt 10
2.303 17. From the unit of K, the reaction is of zero order.
14. rb  K  H  
n
C0  C
k
t
pH = 2  H    102
1.5  0.30
0.08 
n t
r0  K 102 
t  15S
 1
at pH  1,  H   10 2.303 a 2.303 90
18. k  log10 = log10
r1 n t a  x 10  60 90  45
 100  10
n
r1  K 101  r0 (Note that pressure 180mm Hg corresponds for
x(g) unreacted 2y(g) formed and z(g) formed)
n  2
180
15. rate = K[Reactant]m  Pressure due to unreacted x
4
 45 mm Hg
Let concentration of reactant be a mol litre–1
m
 1.15103 sec1 .
then for I order : r1  K1 a  ..... (1) 19. By set I, half-life is 2 min. In set II. Number of
If concentration of reactant be a mol mL–1, then moles have been doubled thus half-life is also
concentration in mol litre1  a 103 doubled, i.el, now it is 4 min. Thus
1
8 moles change to 4 moles in = 4 min and 4

Thus, r1  K1 a 10 3 moles change to 2 moles in = 2 min Thus,
Total time = 6 min
r1  K1 103 a 
0.693
20. t 1 for A  B 
r1  K1' a  ....(2) 2
0.693
Similarly for II order : r2  K 2  106  a 2 t1 0.693
for  C  D 
2 2 0.693  60
r2  K '2 a  ....(3)
3
t 1'
Similarly for III order : r3  K 3 109 a  2

a'
21. For zero order reaction t ''
1 a ''
3
r3  K 3' a  ....(4) 2
 0.693 1 For I order reaction,
22. k  yr when t is 99%
10 0.693 0.693
t1/ 2    1.386sec1
2.303 a 2.303 K 0.5
= log log102
t a  0.99a = t 25. CH3OCH3 g   CH 4  H 2  CO
0.693 2.303 t  0 0.4atm
  log102 or 0 0 0
10 t 99%
t  12 0.4  x x x x
t
10
 2.303  2  66.5 total pressure 0.4  x  x  x  0.4  2 x atm
0.693
2.303 a
years = 70 years K log
t ax
23. Rate = K   A  2.303 0.4
K log
0.04  K  A 10 and 0.03  K  A 20 12 0.4  x
0.693
A 
 10 0.04 4 but K 
  14.5
A  0.03 3
  20
0.693 2.303 0.4
A 
 log
 0.4  x
log  10
2.303 t1 12
Also, t  K A  when t = 10 min 2
  20
x  0.175 atm
2.303 4
10  log
K 3  total pressure= 0.4  2 x  0.4  2  0.175
 0.75 atm
2.303 4
K  log  0.0288min 1 26. Since it is first order reaction, therefore
10 3
0.693 0.693 0.693 0 .6 9 3
t1/ 2    24.06min K
K 0.0288 t1/ 2 = 1 4 . 5
m n
24.  Rate = K  A   B 2.303 a
K log
m
For case I : 0.05  K 0.1 0.2 .... (1)
n t ax
0.693 2.303 100
m
For case II : 0.10  K 0.2  0.2  .... (2)
n
 log
14.3 t 5
m
For case III : 0.05  K 0.1 0.1 ... (3)
n
t  61.82days
m
= 62 days
1 1
By Eqs. (1) and (2), 2   2  0.693
  27. t1/ 2 
K
m=1
0.693 0.693
1
n
28. t1/ 2    1117.7 sec
By Eqs. (1) and (3), 1   2  K 6.2  10 4
 
K2 Ea  1 1 
 n=0 29. log K  2.303R  T  T 
1 0 1  1 2
Thus, Rate = K  A   B  O.R.  1  0  1
    30. K  AeEa / RT when T  , then K  A .
1 0
Now 0.05  K 0.1 0.2 31. A dip in the curve shows the formation of X
(eaction taking place in two steps). Since the
0.05
K   0.5sec1 reaction is exothermic Ea of X must be less.
0.1
 K2 4. Consider the reaction,
32. Given, K  1.75 Cl2 aq   H 2 S aq   S s   2 H aq   2Claq 
1

T1  25C  25  273  298K, The rate equation for this reaction is ,

T2  35C  35  273  308K rate  K Cl2  H 2 S 
(Since temperatur coefficient is ratio of rate which of these mechanisms is/are consistent
constants at 35C and 25C respectively.)
.) with this rate equation ? (AIE - 2010)
K 2 Ea T2  T1  A) Cl2  H 2 S  H   Cl   Cl   HS   slow 
 2.303log10 
K1 R T1T2
Cl   HS   H   Cl   S  fast 
Ea 308  298
 2.303log10 1.75  B) H 2 S  H   HS  ( fast eqilibrium )
1.987 308  298

E a 
2.303  308  298 1.987
 log1.75cal mol1 Cl2  HS   2Cl   H   S  slow 
10 1) (B) only 2) Both (A) and (B)
E a  10.207 kcal mol 1 3) Neither (A) nor (B) 4) (A) only
k E T  T  5. For the non-stoichiometric reaction
33. log k  2.303R  T T 
2 a 2 1

1  1 2  2A  B  C  D , the following kinetic data

were obtained in three separate experiments,
6.9 10 3 Ea  340  300  all at 298 K (AIE-2014 )

2.303  8.31  340  300 
log 5
3.45 10

Ea  112.5 kJ Expt Initial Initial Initial rare of

no Concentration Concentration Formation of
PREVIOUS JEE MAINS (A) (B) (C)
1. The time for half life period of a certain mol lit 1 sec 1
reaction A  products is 1 hour. When the (i) 0.1M 0.1M 1.2  103
initial concentration of the reactant ‘A’ is (ii) 0.1M 0.2M 1.2  103
2.0 mol L1 , how much time does it take for
(ii) 0.2M 0.1M 2.4  103
its concentration to come from 0.50 to
0.25 mol L1 if it is a zero order reaction ?
The rate law for the formation of C is
(AIE - 2010 )
dC dC
 K  A B   K  A  B 
2
1) 4 h 2) 0.5 h 3) 0.25 h 4) 1h 1) 2)
2. The rate of chemical reaction doubles for dt dt
dC dC
every 100 C rise of temperature. If the  K  A B   K  A
2
3) 4)
temperature is raised by 50 0 C , the rate of dt dt
the reaction increases by about (AIE - 2011) 1
A  2 B , rate of disappearance
1) 24 times 2) 32 times 3) 64 times 4) 10 times 6. For a reaction
2
3. A reactant (A) forms two products A  K
B, 1 of ‘A’ is related to the rate of appearance of
Activation energy Ea1 , A  ‘B’ by the expression (AIE-2008 )
K
 C , Activation
2

d  A 1 d  B d  A 1 d  B 
energy Ea2 . If Ea2  2 Ea1 , then K1 and K2 aree 1)   2)  
related as ( AIE -2011) dt 2 dt dt 4 dt

1) K1  2 K 2 e Ea2 / RT 2) K1  K 2 e Ea1 / RT d  A d  B d  A d  B
3)   4)  4
dt dt dt dt
3) K 2  K1e Ea2 / RT 4) K1  AK 2 e Ea1 / RT
 PREVIOUS JEE MAINS - KEY 2. The acid hydrolysis of ester is:
1) 3 2) 2 3) 2 4) 4 5) 4 6) 2 (i) first order reaction
(ii) bimolecular reaction
PREVIOUS JEE MAINS HINTS (iii) unimolecular reaction
x (iv) second order reaction
1. For a zero order reaction K  .....(1) The true statements are
t
1) i, ii 2) All are correct
 A0
K 3) ii, iv 4) ii, iii, iv
for zero order reaction 2t 1 .....(2) 3. Which of the following statements are
2
correct:
since  A0 = 2 M, t 1  1 h , K  1 (i) law of mass action and rate law expressions
2 are same for single step reaction
 from equation (1) (ii)the slowest step of a complex reaction gives
0.25 the order of the complex reaction
t  0.25 h
1 (iii) both order and molecularity have
2. For every 100 C temperature raises rate of normally a maximum value of 3
reaction becomes doubled (iv) molecularity of a complex reaction
3. By based on arrhenius equation A+2B  C is 3
4. Slowest step is the rate determining step. Thus 1) i, ii, iii 2) All are correct
in case (A) rate law is given as 3) ii, iv 4) ii, iii, iv
rate  K Cl2  H 2 S  4. Consider the following reactions at 300 K
While for the reaction given in case (B), rate X  Y  uncatalysed reaction 
1
law is given as rate  K Cl2  H 2 S   H   X 
catalyst
 Y  catalysed reaction 
Hence, only mechanism (A) is consistent with The energy of activation is lowered by
the given rate law
0.314 K J. m ol  1
for the catalysed reaction.
5. The rate only depends upon  A
The rate of reaction is
6. According to rate equation 1) 28 times 2) 15 times 3) 25 times
4) 22 times that of uncatalysed reaction.
LEVEL-IV 5. A substance undergoes first order
decomposition. The decomposition follows
1. For the reaction A  2 B  C  D  2 E the rate two parallel first order reaction as :
equation is rate = K  A  B 0 C  then the rate
is K1 B
i) Doubled when [A] is doubled keeping B K 1  1.26  10  4 sec  1
and C constant A
and K 2  3.8  105 sec1
ii) Doubled when [C] is doubled keeping A
and B constant K2
C
iii)The same when [B] is doubled keeping A
and B constant The percentage distribution of B and C are
iv) Doubled when [B] is doubled keeping A 1) 80% B and 20%C
and C constant
2) 76.83% B and 23.17% C
The correct combination is
1) i, ii, iii 2) All are correct 3) 90% B and 10%C
3) ii, iv 4) ii, iii, iv 4) 60% B and 40% C
6. The rate of the reaction between haemoglobin 1
(Hb) and carbon monoxide (CO) was studied 8. Graph between log k and
T
is a straight line
at 20 0 C . The following data were collected   1
with OX = 5, tan    2.303  . Hence Ea will be
with all concentration units in  mol / L (A  
haemoglobin concentration of 2.21  mol / L
is equal to 2.21 10 6 mol/L )

5
Determine the orders of this reaction 1) 2.303  2 cal 2) cal
2.303
with respect to Hb and CO and rate constant.
3) 2 cal 4) none of these
1)1st order in Hb and Ist order in CO
9.
In a first order reaction the concentration of
0.140 L mol 1 s 1 product 'x' at time 't' is given by the
2) 1st order Hb and 1st order in CO expression (a=initial concentration, k=rate
constant, n=order)
0.280 L  mol 1 s 1
1
1 1
3) 1st order, 2nd order, 0.35 L mol s 1) x  a 
1  e  kt
 2) x
a  x 
4) 2nd, order, 2nd order, 0.24 L mol 1s 1 1 a
3) x  n 1 4) x 
7. Consider following graphs (1) and (2) 2 a  x 
Passage : 1
A collision between reactant molecules must occur
with a certain minimum energy before it is
effective in yielding product molecules. This
minimum energy is called activation energy Ea .
Larger is the value of activation energy, smaller
is the value of rate constant. Larger is the value
of activation energy, greater is the effect of
temperature rise on rate constant K.

Et I
The order of reaction and the value of rate
Potential energy

constant is: Ef
1) First order, 2.37  10 5 min 1 Eb
2) Second order, 2.37 105 torr 1 min 1 R H
1 1 P
3) Zero order, torr min
4) None of the above Collision number

E f  Activation energy of forward reaction

Eb  Activation energy of backward reaction
H  E f  Eb
Et  Threshold energy
Answer the following questions: Passage : 2
1. If a reaction, A  B  C , is exothermic to The energy profile diagram for the reaction
the extent of 30 kJ/mol and the forward CO  g   NO2  g   CO2  g   NO  g  is
reaction has an activation energy of 249
given below
kJ/ mol, the activation enegy for reverse reaction
in kJ/mol is
I
1) 324 2) 279 3) 40 4) 100
2. For the following reaction at a particular

Energy
temperature, according to the equations,
2 N 2O5  4 NO2  O2
R
P
1
2 NO2  O2  N 2O5 the activation energies Reaction path
2
are E1 and E2 respectively. Then Answer the following questions:
1) E1  E2 2) E1  E2 1. The activation energy of the forward reaction is
1) x 2) y 3) x  y 4) x  y
3) E1  2E2 4) E1 E22  1 2. The activation energy of the backward reaction
3. In a hypothetical reaction, A  Y , the activation is
energies for the forward and backward reactions 1) x 2) y 3) x  y 4) x  y
3. The heat of the reaction is
are 15 and 9kJ mol 1 respectively. The potential
1
1) x 2) y 3) x  y 4) x  y
energy of A is 10kJ mol . Which of the 4. The threshold energy of the reaction is
following is wrong ? 1) x  y  z 2) x  y  z
1) Threshold energy of the reaction is 25 kJ
3) x  y  z 4) x  y  z
2) The potential energy of B is 16 kJ
3) Heat of reaction is 6 kJ
4) The reaction is exothermic LEVEL-IV - KEY
1) 1 2) 1 3) 1 4)1 5)2 6) 2 7) 2
4. For two reactions, activation energies are Ea1
8) 3 9) 1
and Ea2 ; rate constants are K1 and K2 at the same Passage - I:
1) 2 2) 1 3) 4 4) 3 5) 4 6) 1
temperature. If K1  K2 , then
Passage - II:
1) Ea1  Ea2 2) Ea1  Ea2 1) 1 2) 3 3) 2 4) 3
3) Ea1  Ea2 4) Ea1  Ea2
LEVEL-IV - HINTS
5. The rate constant of a certain reaction is given
r  K  A B  C 
0
by K  Ae Ea / RT ( where A = Arrhenius constant 1.
). Which factor should be lowered so that the 2. In acid hydrolysis of ester. water is taken in
rate of reaction may increase ? excess quantity r  k  Ester 
1) T 2) Z 3) A 4) Ea order = 1
6. The activation energies for forward and molecularity = 2
backward reactions in a chemical reaction are 3. Molecularity is always a whole number
30.5 and 45.4 kJmol 1 respectively. Then 4. K uncat  A e Euncat / RT
reaction is
K Cat  A e Ecat / RT
1) exothermic 2) endothermic
3) neither exothermic nor endothermic
4) independent of temperature
 Let the Ea for uncatalysed reaction be A, then
for catalysed reaction it will be  A  8.314  103 
calaries/mol
 A 8.314103  / RT
K cat e 
 e8.31410 / RT  e3.3  28
3
  A / RT
K uncat e
5.. For parallel path reaction
K average  K1  K 2  1.26 104  3.8 105
 1.64 104 sec1
Also, Fractional yield of B
K B 1.26 104
= K   0.7683
av 1.64 104
Fractional yield of A
K A 3.8 105
= K   0.2317
av 1.64 104
6. Compare that 1 and 2 in the table. It is clear
when the concentration of Hb is double, the rate
is also doubled, hence it is first order with
respect to Hb, further an examination of data 2
and 3rd shows that it is also of first order with
CO.
Therefore
Rate of reaction
= [ Hb]1[CO]1  Rate constant
0.619 mol L s 1
= 2.2 mol / L 1.00 mol / L  Rate constant
0.619
 Rate constant =
2.2N
= 2.280 L mol 1s 1
1 1
7.  kt 
p p0
Ea
8. log k = log A 
2.303RT

Slope =
Ea

1
2.303R 2.303
 given 
E a  2.303R  slope  R  2cal.
9. According to arrhenius equation.