View - aisect

AISECT TUTORIALS : CHEMISTRY : SET-6
CHAPTER-1
CHEMICAL KINETICS
1. Chemical Kinetics
The study of the rate of reaction, the
mechanism by which the reaction takes place
and the influence of temperature and
pressure on the rate of reaction is known as
chemical kinetics Some reactions are
spontaneous or vigorous while some
reactions are very slow kietics of each of
reaction cannot be studied and therefore,
they are beyond the scope of kinetics.
1.1 Reaction Rate or Rate of Chemial
Reaction
The speed or rate of reaction is defined as
the rate of disappearance of a starting
reactant or the rate of apperance of a
product in a unit time.
For the reaction A → B
Rate of reaction= Rate of disappearance of A
(Reactant)
R ∝ − dA
dt
The minus sign indicates that the
concentration of A Raditant decreares with
time.
Rate of reaction = Rate of appearance. of B
(Product )
R ∝+
dA
dt
The rate of reaction at any instant is
influenced by the concentration of reacting
substances and is given by law of mass
action. Consider a reaction.
A → Products
Suppose dC be the small concentration of the
substance. A undergoing change in time dt.
The rate of the reaction i.e., dC/dt will be
proportional to C A
or [A], i.e. molar
concentration of A. Now if we start with a
gm moles of substance A and let x gm moles
undergo change in time t. Then A →
Products
a 0 Original
(a-x) x After time t
Now applying law of mass action, we have
dC
∝ (a − x) or dC = k(a − x)
dt dt
Where k is known as the rate constant of
the reaction. When molar concentration (a-x)
is unity, we have
dC
= k
dt
Thus at a fixed temperature, rate of reaction
is equal to the rate constant when the
molar concentration is unity.Since the
concentration is expressed in moles per
litre, and time in seconds the unit of rate of
reaction is moles. lit -1 . sec -1 .
1.2 Average Rate of Reaction
The rate can be determined experimentally
by following the change of concentration of
any of the reactants or products with time.The
average may then be defined by the equation.
Average rate of reaction=
Amount of reactants consumed Amountofproductformed
=
Time interval
Timeinterval
(1)

AISECT TUTORIALS : CHEMISTRY : SET-6
1.3 Rate Constant
The rate of reaction when concentration of
the reactant is 1 mole/litre. It is a constant for
a particular reaction at a given temperature
and also known as specific rate constant or
specific reaction rate.
Velocity constant is defined as the velocity
of the reaction if the molar concentration is
unity at fixed temperature.
Factors Influencing the rate of reaction
The rate of reaction depends upon various
factors, which are given below
(a) Nature of Reactants - Each reaction has
its own characteristic activation energy. In a
reaction with higher energy barrier the
reactant molecules require higher energy of
activation of pass the barrier and get
converted into products.The number of
molecules possessing this amount of
energy will therefore be less and hence
the rate of reaction is slow. In a reaction ,
with lower energy barrier the reactant
molecules require relatively small energy of
activation to pass the energy barrier and
hence the number of molecules
processing the required energy will be
more. As a result rate of reaction will be
fast. This indicates that rate of reaction
depends upon the nature of reactants. How
rate of reaction depends upon the nature of
reactants may be made clear by the fact that
iron undergoes rusting in presence of moist
air, copper gets tranished in air, but gold
remains unaffected. Similarly, ferrous
sulphate solution quickly decolourise the
acidifeid solution of KMnO 4
but oxalic acid
decolourises it very slowly at room
temperature.
(b) Concentration - With the exception of
certain zero order reactions, upon which the
concentration is without effect, an increase
in the concentration of reactants results in
the acceleration of reaction rate. For
example, if we take 5 mL portions of 0.05 M
0.1 M 0.25 M, 0.5 M 1.0 M and 1.5 M
solutions of sodium thiosulphate and add 5
mL portions of 0.5 M HCl in each solution
then sulphur will be precipitated first in the 1.5
M solution of sodium thiosulphate and if we
plot a graph between original concentrations
of sodium thiosulphate and reciprocals of
time,it will be found that Rate of reaction ∝
Reciprocal of time
sodium thiosulphate.
∝
Concentration of
Similarly if in a reaction X+Y → XY, if the
concentration of X and Y are doubled the rate
of reaction will increase to 4 times and if the
concentration of both X and Y are increased
by three times, the rate of reaction wil
increase to nine times. The rate of reaction is
influenced by the concentration is also
evident from the fact that zinc reacts rapidly
with Conc. HCl but only slowly with dilute HCl.
Carbon burns in oxygen much more quickly
than in air. An increase in concentration
increases the number of molecules per unit
volume. As a result, number of collisions
increases and greater number of molecules
acquire extra energy required for the reaction.
Hence rate of reaction increases with
increase in concentration.
(c) Pressure - Pressure has very little effect
on rates of reactions involving solids and
liquids.For reactions involving gases
however, an increase in pressure
increases the concentration by bringing the
(2)

AISECT TUTORIALS : CHEMISTRY : SET-6
molecules close together and hence increase
the probability of them colliding (hence
increase the rate of rection). Pressure does
not alter the value of rate constant.
(d) Subdivision :- The rate of reaction is
increased if the reactants are in subdivided
form. According to collision theory, the rate of
a reaction depends upon the number of
molecular collisions between reactants, not
every collision leading to reaction. When a
substance is in finely divided form its
surface area is much increased and so
more atoms or molecules are allowed to
collide. Larger the surface area, higher
will be the rate of reaction. Hence rate of
reaction increases with subdivision. For
example if we take two breakers containing
similar volumes of dilute HCl in both of them
and then place a small lump of limestone in
one beaker and the same qunatity of
limestone powder in the other beaker, we will
find that rapid action takes place in the
beaker in which limestone is added in the
powdered form. This indicates the effect of
greater surface areas of the limestone
powder.
CaCO 3
+ 2 HCl → CaCl 2
+ CO 2
+ H 2
O
(e) Temperature An increase in temperature
causes the molecules to travel faster and
hence increases the probability of them
colliding. The rate of reaction thus
frequently increases with the rise in
temperature. In most cases, a rise in 10 0 C in
temperature doubles the rate of reaction.
This means that foods should cook twice
as fast in a pressure cooker at 110 o C as in
an open saucerpan and deteriorate four
times as rapidly at room temperature (25 0 )
as they do in a refrigerator at 5 0 C. The ratio
of the rate constants of a reaction at two
temperatures separated by 10 0 C (usually
25 0 C and 35 0 C) is known as temperature
coeffcient. Thus Temperature Cofficient=
K 35
/K 25
=2 or 3
The temperature cofficient for most chemical
reactions lies between 2 and 3. Few reactions
(e.g., reaction between NO and O 2
to form
(NO 2
) exhibit a small negative temperature
cofficient and so the rate of such reactions
decreases with rise in temperature. Increase
in temperature increases the number of
activated molecules to a much greater extent
than the number having average kinetic
energy.
The rate of a reaction doubles for every
10 0 C rise in temperature from 298 K to
308 K probably due to the fact that
effective collisions double for 10 0 C rise
in temperature from 298 K to 308 K.
Since energy of activation of majority of
reactions lies in the range 50-55 kJ
mole -1 therefore the rate of reaction
doubles for all such reactions when the
temperature is incresed from 298 K to
308 K. Only those reactions whose
activation energy lies in the range 50-55
kJ are found to double their rate for this
range of temperature.
The velocity of motion of molecules
increases with temperature and they
collide at a higher frequency. Moreover
the molecules becomes more active at
higher temperatures, as a result of which
the number of effective collisions
increases. These are the causes by
which the rate of reaction increases with
(3)

AISECT TUTORIALS : CHEMISTRY : SET-6
temperature. Experience show that
when temperature increases 10 0 C, the
reaction rate increases approximately
2-3 times.
The ratio of the velocity constants for
two different temperatures (generaly
25 0 C and 35 0 C) is called the
temperature coefficient of a chemical
reaction rate.
Activation energy is the minimum excess
energy, compared with the mean energy
of the reacting molecules at a given
temperature, which the molecules must
possess if their collision is to produce a
new product.
A chemical process is caused by the
collision of molecules having energies
i.e.active molecules. The energy
required to activate the initial particles is
called the activation energy of the
reaction. In some cases the activation
energy is the main factor determining the
rate of a chemical process. The higher
the energy, the less the number of
molecules possessing this energy at a
given temperature and the slower the
chemical reactions. It has been
established that process with activation
energies less than 10 kcal/mole proceed
at a high rate of ordinary temperatures,
while the rate is extermely low at
activation energies over 30 k cal/mole
The less the energy of activation for a
reaction the more easily that reaction will
take place, for example for a reaction with
energy of activation 20 k.cals per mole the
rate of reaction would be quite good even at
ordinary temperatures but for a reaction with
energy of activation 40 k.cals ner mole, the
reaction will proceed at a reasonable rate
provided the temperature is raised to about
400-500 0 C.,
As the temperature increases, the
average, K.E. also increases and a
greater fraction of molecules, have
higher kinetic energies. As a result of
increase in average kinetic energy of the
reacting species with increse in
temperature, a greater fraction of
collisions leads to reaction at higher
temperature. The frequency of collisions
also increases with increase in
temperature. The net result of greater
frequency of collision and larger fraction
of collisions having energies above the
activation energy lead to a rapid
increase in the rate of reaction with an
increase in temperature.
The reaction between CO and NO 2
from CO 2
and NO at about 200 0 C is an excellent
example, which shows the effect of
temperature on the rate of reaction. When
the mixture of CO and NO 2
(reddish brown
gas) is heated at about 200 0 C, the reddish
brown colour of NO 2
gas disappears slowly
forming colourless CO 2
and NO gas. But as
the temperature is increased form 200 0 C to
say 250 0 C the colour disappears more readily
indicating that the rate of reaction has
ncreased by increasing the temperature of
the reaction
CO(Colourless) + NO 2
(Reddish brown) →
CO 2
(Colourless) + NO (Colourless).
Reactions generally proceed more readily
when the products have a lower energy
content than the reactants, the difference
(4)

AISECT TUTORIALS : CHEMISTRY : SET-6
being the free energy of reaction. Highly
endothermic reactions are unlikely, to take
place readily at low temperatures. The energy
of activation has to be greater than the heat
of reaction so that activation energy for any
endothermic reaction with a high heat of
reaction must be high. In case of exothermic
reactions, i.e. E P
-E R
=-∆E.
For an exothermic reaction, the energy of the
reactants is higher than that of the products
and energy is released when products are
formed. But before the products can be
formed, the system must acquire additional
energy and it must pass through the activated
state in which the potential energy is
maximum. The energy that the reactants
must acquire to reach the activated
comples is the activation energy of the
forward reaction. The activation energy of
the backward reaction (E backward
) should
therefore be greater than the activation
energy of the forward reaction (E forward)
. For
exothermic reaction.
- ∆H = (E backward
) > (E forward
).
∆H is negative as it must be for
exothermic reaction.
For an endothermic reaction, the reactants
are at lower potential energy than the
products and the activated state is at higher
energy than either the reactants or the
products. The equation for endothermic
reaction is ∆H = (E backward
) < (E forward
).
A reaction with large activation energy is
said to have a high potential energy
barrier. Such type of reactions have
slower rates of reaction than the reactions
with low potential energy barriers (that is
reactions for which activation energy is
small.
(f) Catalyst - A catalyst is a substance which
lowers the energy of activation of a reaction.
We know that reactants in a chemical
reaction have to cross an energy barrier
before they can react to form the products.
Thus higher the energy barrier, the slower is
the rate of reaction. A catalyst therefore
functions by offering an alternative route for
the reaction which involves a lower energy of
activation. The magnitude of energy barrier is
reduced in presence of a catalyst and hence
a large number of particles of reactants can
get over it and as a result the rate of
reaction increases.
Since the lower energy barrier applies to
both forward and backward reaction of a
reversible reaction, the final equilibrium
positiion can not be affected in any way.
Thus a catalyst speeds up both the
forward and reverse reaction by the same
amount and does not alter the position of
equilibrium in a reversible reaction.
Heterogeneous catalysts allows molecules
to reside on their surface in such an exposed
condition that the majority of collisions bring
about reactions. Homogeneous catalysts
allow collisions to take place in such a
manner that two or more simple reactions are
subsituted for the complicated one which
needs a lot of energy to occur.
2. Molecularity of Reaction.
The total number of molecules present in the
reactant(s) of a balanced equation is known
as molecularity of the reaction. For example,
PCl 5
→ PCl 3
+ Cl 2
(Unimolecular)
(5)

AISECT TUTORIALS : CHEMISTRY : SET-6
2HI →H 2
+ l 2
(Biomolecular)
CH 3
COOC 2
H 5
+H 2
O ⇌ CH 3
COOH +C 2
H 5
OH
(Bimolecular)
(i)
(ii)
(iii)
Molecularity of a reaction is always a
whole number.
Molecularity of areaction is never zero
Reactions having molecularity of more
than three are rare. It is because the
chances of simultaneous collisions
between three or more particles are
rare.
3. Order of Reaction
It is defined as the number of molecules
whose concentration (changes) determines
the rate of reaction. In other words, it is the
sum of the powers of the concentration of
reactants in the rate equation (rate law).
Consider a reaction 2 NO + O 2
→ 2NO 2
As determined from rate law.
Rate = K [NO] 2 [O 2
]
∴Order of reaction with respect to NO is 2
Order of reaction with respect to O 2
is 1
The overall order of reaction is 2+1 = 3
Examples of first second and third order
reactions.
slow
(i) H 2
O 2 ⎯⎯→ H 2
O + O
fast
O + O ⎯→ O 2
Since the slow (rate determining) step
involves only one molecule the order of
reaction is 1 and not 2 although reaction is
usually written as
2 H 2
O 2
→ 2 H 2
O + O 2
(ii) Similarly
2N 2
O 5
→4 NO 2
+ O 2
(1st order)
(iii)
COOH
COOH
→ CO + CO 2 + H 2 O (1st order)
(iv) CH 3
COOC 2
H 5
+H 2
0⇌CH 3
COOH +C 2
H 5
OH
(1st Order)
(v) C 12
H 22
O 11
+ H 2
O→H 2
O
C 6
H 12
O 6
+ C 6
H 12
O 6
(1st order)
Glucose Fructose
(vi) (CH 3
CO) 2
O + 2C 2
H 5
OH →
Acetic anhydride Ethanol
2 CH 3
COOC 2
H 5
+ H 2
O (1st order)
Ethyl acetate
(vii) 2HI →H 2
+ l 2
(2nd order)
(viii) CH 3
COOC 2
H 5
+NaOH →
CH 3
COONa + C 2
H 5
OH (2nd order)
(ix) 2 NO + O 2
→ 2 NO 2
( 3rd order)
(x) 4 KClO 3
= 3 KCIO 4
+ KCI (4th order)
Remember that
(a)
(b)
(c)
Order of reaction is an experimentally
determined quantity.
Order of reaction cannot be written from
the balanced chemical equation.
Molecularity and order of a reactions
may be same or different.
4. Pseudo-Unimolecular Rections.
Although in most of reactions order and
molecularity are same, there are certain
reactions whose order and molecularity differ
For example hydrolysis of sugar cane.
C 12
H 22
O 11
+ H 2
O → C 6
H 12
O 6
+ C 6
H 12
O 6
Sucrose Gluose Fructose
Molecularity of this reaction is 2 but its order
is 1 because its rate depends only on the
concentration of surcose, the concentration of
(6)

AISECT TUTORIALS : CHEMISTRY : SET-6
water is very high and does not change
during the reaction (i.e. concentration of
water remains practically constant throughout
the reaction). Such reactions are known as
psendounimolecular or pseudo first order
reactions. Other example of
pseudounimolecular reaction is the acidic
hydrolysis of esters where water remains in
excess.
H + + CH 3
COOC 2
H 5
+ H 2
O →CH 3
COOH +
C 2
H 5
OH + H +
Although it is termolecular (molecularity =3)
reaction its order is one as concentration of
H + and H 2
O remain constant during reaction.
Hydrolysis of organic chlorides is also an
example of first order reaction.
RCI + H 2
O → ROH + HCI
since water (one of reactants) is again in
large excess and its concentration remains
constant throughout the reactions
Thus when one of the reactants is present in
large excess, the second order reaction
conforms to the first order and is known as a
pseudounimolecular reaction.
Reaction between acetic anhydride and
excess of ethanol to form ester and
conversion of N-Chloroacetanilide to
p-chloroacertanilide are also examples of
pseudounimolecular reactions.
Further although reactions may have zero or
a fractional order (e.g. 1/2, 3/2 etc.) the
molecularity must always be an integer and
never zero.
On considerng the following reactions
5.
TABLE DIFFERENCE BETWEEN ORDER AND MOLECULARITY OF A REACTION
Order of Reaction
1 1. It is the sum of the powers of concentration
terms of the reactants in the rate equation
2 It may be whole number, zero and even
fraction.
3 It is to be determined from the experiment and
depends upon the experimentally determined
rate of the over all reaction
4 Order of reaction is the same for the whole
reaction, no matter it is simple or complex
reation
5 It refers to a reaction as a whole irrespective of
the number of steps involved and need not
necessarily be related to stoichiometric
equation of the reaction.
Molecularity of Reaction
1 1. It is the number of molecules or ions of the
reactants taking part in a single step (rate
determining step)
2 It is a whole number and is never zero.
3 It is concerned with the reaction mechanism
and purely a theoretical value obtained from
the balanced single step reaction.
4 Molecularity in a multistep reaction is
expressed for each step
5 It depends upon the rate determining step in
the reaction mechanism.
(7)

AISECT TUTORIALS : CHEMISTRY : SET-6
(a) A → Product, Rate
∝
C A
or [A]
(b) A+A → Product, Rate ∝ C A
C A
Or [A] 2
(c) A + A + B → Product Rat
(d) A+A+B→ Product Rate
∝
∝
C A
C B
Or [A] [B]
C A
C A
C B
Or [A 2 ] [B]
(e) A+A+B→Product Rate ∝ C A
C B
C B
Or [A] [B 2 ]
It is seen that rate depends on different
concentrations. Thus order of reaction is also
defined as the number of molecules whose
concentration alters as a result of chemical
change.
Molecularity of a reaction is theoritical value
and order of reaction is practial value.
Sometimes the order o the reaction differs
from the molecularity of the reaction as given
by the stoichiometric equation.
6. Zero Order Reactions.
Reactions whose rate is not affected by
concentration or in which the concentrations
of the reactants do not change with time are
called zero order reactions. Thus the rate of
such reactions remains constant.
Rate = K
Many photochemical reactions (e.g. formation
of HCl from H 2
and Cl 2
) and some
heterogeneous reaction (e.g. decomposition
of
hydrogen iodide and ammonia on the
surfaces of gold and tungsten) are the
examples of zero order reactions.
sun light
H 2
(g) + Cl 2
(g) → 2 HCl (g)
Characteristics. (i) The rate of reaction is
independent of the concentration of the
reacting substance. Concentration of
products increases lineraly will time. Plot of
concentration of products with time is a
straight line passing through origin.
(ii) The unit of zero order rate constant is
mole litre -1 time -1
(iii) The half life is directly proportional to the
initial concentration of the reactants.
7. First Order Reactions.
Reactions whose rate is determined by the
change of one concentration term only are
known as first order reactions.
Consider a general reaction of the first order
A → Products
The rate of such reaction at any moment will
thus be given by the expression
− dcA ∝ [C A ]
dt
or − dc[A] = K CA
dt
or − dc
dt = K [A]
Where C A
is the concentration of the reactant
A at the moment when the rate of reaction is
determined and K is rate constant, specific
rate constant or velocity constant.
7.1 Graphical Method of
Determination of first order reaction
K = 2.303
t
log ⎛ a
⎝ a − x ⎞ ⎠
or
Kt
= log a − log (a − x)
2.303
Kt
or log (a-x) = log a -
2.303
This is an equation of a straight line. So if a
graph of log (a-x) against 't' is plotted, it will
K
be a straight line with the slope = - and
2.303
intercept = log a.
(8)

AISECT TUTORIALS : CHEMISTRY : SET-6
Fig. 1
Sometimes there is an uncertainity about the
initial concentration because the instant at
which the reaction begins is not exactly
known. In such a case, the value of 'a' may
be eliminated by taking concentrations (a-x 1
)
and (a-x 2
) at time intervals t 1
and t 2
respectively.
2.303
Now K= log −
t
t= 2.303
K log a
(a − x)
. . . t 1
= 2.303
K log a
(a − x 1 )
a
(a − x)
and t 2
= 2.303
K log a
(a − x 2 )
. . . t 2
-t 1
= 2.303
K log (a − x 1)
(a − x 2 )
and K = 2.303
t 2 − t 1
log (a − x 1)
(a − x 2 )
7.2 Examples of first Order Reaction
or
(1) Hydrolysis of an ester by acid
CH 3
COOCH 3
+ H 2
O →CH 3
COOH + CH 3
OH
It is an example of pseudo unimolecular
reaction. Water is present in large excess
and therefore its concentration remains
constant, only the concentration of methyl
acetate changes. Hence the rate of reaction
is determined by the concentration of methyl
acetate only. So that the reaction is a first
order reaction
(2) Inversion of sugar- Cane sugar is
hydrolysed in presence of mineral acids as
catalyst.
[H
C 12
H 22
O 11
+ H 2
O + ]
→ C 6
H 12
O 6
+ C 6
H 12
O 6
cane sugar glucose fructose
The reaction is bimolecular but it is first order
reaction. This is also an example of pseudo
unimolecular reaction.
(3) Decomposition of hydrogen peroxide
H 2
O 2
→ H 2
O + O (slow)
O+O →O 2
(fast)
dx
dt
= K[H 2 O 2 ]
(4) Decomposition of N 2
O 5
N 2
O 5
→N 2
O 4
+
1
2
N 2
O 4
→ 2 NO 2
(fast)
dx
dt
= K[N 2 O 5 ]
O 2
(slow)
(5) Decomposition of sulphuryl chloride
SO 2
Cl 2
→SO 2
+ Cl 2
(slow)
dx
dt
= K[SO 2 Cl 2 ]
(6) Radioactive disintegration
R A
→R B
+ α-particles
(7) Decomposition of ammonium nitrite
NH 4
NO 2
(aq)
∆
→⎯
Ammoinum nitrite
Dinitrogen
2H 2
O (l) + N 2
(g)
water
Characteristics of first order reactions.
1. Rate of reaction. The rate of reaction is
directly proportional to the concentration of
the reacting substance
2. First order rate constant. It is a
characteristic constant of a particular reaction
at a given temperature. It does not depend
upon initial concentration of the reactants,
time of reaction and extent of reaction. Its unit
(9)

AISECT TUTORIALS : CHEMISTRY : SET-6
is time -1 , i.e. if it is expressed in seconds, K
is expressed in seconds -1 , if it is expressed in
minutes, K is expressed in minutes -1 .The
value of K does not change with
concentration units because a/(a-x) will be
same whatever be the units of concentration.
3. A plot of log a/(a-x) versus time is linear
passing through origin with slope = - K/2.303
4. Half life period (Half life time t 1/2
,). Half life
of a reaction is the time required to convert
the original concentration of reactant to half.
For first order reaction, at half time i.e., at t 1/2
,
xbecomes a/2.
Therefore, putting t=t 1/2
and x=a/2 in eq. (i) we
get
K = 2.303
t1/2 log 2
or t 1/2
= 0.693
K
Note that half life time of a first order
reaction is constant and independent of the
initial concentration of the reactant.
If the quantity of a reactant at start(i.e., when
t=0) is a o
and the quantity that remains after n
half life time is a n
, then
a n = ⎡ ⎣
1
2 ⎤ ⎦ n x a 0
Remember that all radioactive decays are
examples of first order reactions.
8. Second Order Reactions.
Reactions whose rate is determined by
change of two concentration terms are known
as second order reactions. For example,
For a general reaction
A + B → products
dx
dt
= K[A] 2 [B] 0
or dx
dt
= K[A] 0 [B] 2
dx
dt
= K[A] [B]
Thus the rate of a second order reaction
varies directly as the square of the
concentration of reactant.
Characteristics. (i) Rate of reaction is
directly proportional to the square of the
concentration of the reacting substance.
(ii)
(iii)
(iv)
(v)
The unit of second order rate constant
is litre mole -1 time -1 . The value of K
depends upon the unit in which
concentration of the reactant(s) is
expressed.
The half life of a second order reaction
is inversely proportional to the initial
concentration of the reactants and rate
constant (cf. half life period of a first
order reactions is inversely proportional
to only K an independent of a).
All the second order reactions obey the
following kinetic equation.
When a graph is plotted between t and
1 /(a-x) a straight line is obtained; the
slope of the line gives 1/K.
8.1 Examples of second order
Reaction
1. Hydrolysis of an ester by an alkali
CH 3
COOC 2
H 5
+ NaOH → CH 3
COONa +
C 2
H 5
OH
In this case concentration of ester and base
are changed during the hydrolysis process,
so it is a second order reaction.
(2) Conversion of a ammonium
cyanate into urea
It is a slow process. It occurs as follows
NH 4
CNO⇌NH 4
NCO⇌HN=C=O+NH 3
(fast)
NH 2
H-N=C=O + NH 3
→ O=C < HN 2
(slow)
(10)

AISECT TUTORIALS : CHEMISTRY : SET-6
Hence, the slowest step determines the rate,
so it is a second order reaction.
(3) Benzoin condensation
C 6
H 5
CHO+ OHCC 6
H 5
Benzaldehyde
COC 6
H 5
KCN
−−−→
Catalyst
C 6
H 5
CH(OH)
Benzoin
(4) Conversion of ozone into oxygen at 373 K
2O 3
→ 3O 2
(5) Dissociation of HI at 829 K
2HI → H 2
+ l 2
(6) Thermal decomposition of chlorine
monoxide
2Cl 2
O → 2Cl 2
+ O 2
(7) Interaction of alkyl halides with tertiary
amines or pyridine
C 2
H 5
l + C 6
H 5
N(CH 3
) 2
→C 6
H + N(CH ) C 5 3 2 2H + 5 + 1 −
(8) Pseudo-bimolecular reaction- The
reduction of bromic acid to hydrobromic acid
in presence of HI which has molarity seven.
This reaction is also second order.
HBrO 3
+ 6 HI →HBr + 3H 2
O + 3l 2
9. Third Order Reaction
A reaction is of third order if the number of
molecules whose concentration alters as a
result of chemical change is three and this
may be any of the three types
(i) A + A + A → products or 3 A →products
dx =K [A] 3
dt
(ii) A+A+B→products or 2A +B →products
dx
= K [A] 2 [B]
dt
(iii) A + B +C → products
dx
= K [A] [B] [C]
dt
The rate expression varies with concentration
of reactants. When the concentration of all
the three reactants is same, the expression
arise is
K 3
= 1 t
x(2a−x)
2a 2 (a−x) 2
9.1Characteristics of Third Order Reaction
(i) Unit of third order rate constant- The unit is
sec -1 litre 2 mol -2
(ii) The time for completioin of any definite
fraction of reaction is inversely proportional to
the square of initial concentration of the
reactant or t ∝ 1 a 2
(iii) Change in concentration alters the value
of K since the concentration terms in
denominator and numerator are not same in
number.
9.2 Examples of Third Order Reaction
(1) Reduction of ferric chloride by stannous
chloride 2fecl 3
+ Sncl 2
→ 2Fecl 2
+ Sncl 4
(2) The interaction of sodium formate and
silver acetate
2 CH 3
COOAg + HCOONa → CH 3
COOH +
CH 3
COONa + CO 2
+2Ag
(3) Reaction of nitric oxide and hydrogen
2NO + 2H 2
→N 2
+ 2H 2
O
This reaction is of third order although its
molecularity is four.
(4) Combination of ozone by nitric oxide.
O 3
+ 3NO →3NO 2
(5) Reaction of nitric oxide with chlorine,
bromine and oxygen
2NO + Cl 2
→2NOCl
2 NO + Br 2
→2 NOBr
2No + O 2
→2NO 2
(11)

AISECT TUTORIALS : CHEMISTRY : SET-6
Units of Reaction Rate Constants
If concentration represents in mol litre -1 and
time in second -1 , the units of reaction rate
constants are as follows
(a) For first order reaction K = Sec -1
(b) For second order reaction K = litre mol -1 Sec -1
(c) For third order reaction K = litre mol -2 Sec -1
(d) For zero order reaction K = mol litre -1 Sec -1
Rate equations of various order
Order Differntial form Integrated form
0
1
2
3
dx
dt
= K
K = X t
dx
= K(a − x) K = 1 dt t log a
(a−x)
dx
dt
= K(a − x) 2
dx
dt
= K(a − x) 3
K = 1 t
K = 1 2t
x
a(a−x)
x(2a−x)
a 2 (a−x) 2
The expression which shows constancy of
value of K indicates the appropriate order of
the reaction.
10. Symbolic Rate Expression for the
Reaction
2N 2
O 5
→ 2N 2
O 4
+ O 2
Rate of reaction is written as follows
=− 1 d(N 2 O 5 )
= + 1 d(N 2 O 4 )
2 dt 2 dt
= + d(O 2)
dt
Order of Reaction for a General Reaction
aA + bB +cC + .......→ products
Rate = K[A] 1 [B] m [C] n
Where l=order of reaction with respect to A
m=order of reaction with respect to B
n= order of reaction with respect to C
Overall order of the reaction =l+m+n
Illustration 1. Decompositioin of N 2
O 5
occurs
in the following way.
2N 2
O 5
4NO 2
+ O 2
Write the different wasy in which the rate of
reaction can be expressed. Give the relation
between the different rate constants.
Solution The rate of decompositon of N 2
O 5
can be expressed in the following three ways
− d[N 2O 5 ]
= K [N 2 O 5 ]
dt
or − d[NO 2]
= K 1 [N 2 O 5 ]
dt
or − d[O 2]
= K 2 [N 2 O 5 ]
dt
From the equation, it is clear that 1 mole of
N 2
O 5
on decomposition gives 2 moles of NO 2
1
and mole of O
2
2
... K 1
= 2K and K 2
= 1 2 K
Thus it is very essential to specify with
respect to which substance the rate of
reaction is expressed.
Illustration 2. The rate constant for the
forward and backward reactions of hydrolysis
of ester are 1.1 x 10 -2 and 1.5 x 10 -3 per
minute respectively.
CH 3
COOC 2
H 5
+H+⇌CH 3
COOH+C 2
H 5
OH
Calculate the equilibrium constant of the
reaction.
Solution. Given K f
= 1.1 x10 -2 per minute
K b
= 1.5 x 10 -3 per minute
Equilibrium constant
K f
K b
1.1 x 10−2
1.5 x 10 −3
K = = = 7.33
Illustration 3 The ionisation constant of
NH + 4
in water is 5.6 x 10 -10 at 25 0 C. The rate
constant for the reaction of NH + 4
and OH - to
form NH 3
and H 2
O at 25 0 C is 3.4 x 10 10 L
mol -1 s -1 . Calculate the rate constant for
proton transfer from water to NH 3
Solution
[I.I.T.]
(12)

AISECT TUTORIALS : CHEMISTRY : SET-6
NH 4
+ ⇌NH 3
+ H + ; K= 5.6 x 10 -10 ......(i)
Also
H 2
O⇌OH - + H + ; K 2
= 1 x10 -14 .......(ii)
K= K1 5.6 x 10−10
K 2
= = 5.6 x 10 4
10 −14 +
Subtracting eq. (ii) from eq. (i) we get NH 4
OH⇌NH 3
+ H 2
O;
K f
=3.4 x 10 10 L mol -1 s -1 (given)
∴ NH 3
+ H 2
O = NH 4 + + OH - ;
Thus K= K f
K b
Kb= ?
Kf
or K b
= =
3.4 x 1010
= 6.07 x10 5
K 5.6 x 10 4
Illustration 4. The kinetics of the reaction
mA + nB + pC → m'X + n'Y + p'Z obey the
rate expression
dx
dt
= K[A] m [B] n
Calculate : (i) Order of reaction with respect
to A.
(ii) Order of reaction with respect to B.
(iii) Order of reaction with respect to C.
(iv) Total order of reaction.
(v) Molecularity of the reaction.
(vi) Order of reaction if B is taken in large
excess.
Solution. From the rate expression it is
obvious that
(i)
(ii)
(iii)
the reaction is of mth order with respect
ot A
the reaction is of nth order with respect
ot B
the reaction is of zero order with respect
to C because C does not appear in the
rate expression i.e. the rate of reaction
does not depend upon the
concentration of C.
(iv) the total orde of reaction :
(v)
m+n+ zero = m+n
The molecularity of reaction; m+n+P
(vi) When B is taken in large excess, the rate
becomes independent of B and thus
under such circumstance, the rate of
reaction will be determined only with
respect to A. Hence the order of
reaction will be m.
Illustration 5. The reaction 2A + B +C → D +E
is found to be first orde in A, second in B and
zero order in C.
(i)
(ii)
Give the rate law for the above reaction
in the form of differntial equation.
What is the effect on the rate of
increasing concentration of A,B and C
two times ?
(Roorkee)
Solution (i) The rate law for the reaction is
given by
r = dx
dt = K(A) (B)2 (C) 0
r = dx = K(A) (B)2
dt
(ii) On increasing the concentration of A, B,
and C two times.
r 1 = dx = K(2A (2B) 2 (2C) 0 = 8K (A) (B) 2
dt
Thus rate increases eight times.
Illustration 6. Rate of reaction, A + B →
Products is given below as a function of
different initial concentration of A and B.
(13)

AISECT TUTORIALS : CHEMISTRY : SET-6
[A] in mol
litre -1
[B] in mol
litre -1
Initial rate in mole
litre -1 min -1
(i) 0.010 0.010 0.005
(ii) 0.020 0.010 0.010
(iii) 0.010 0.020 0.005
Determine the order of reaction with respect
of A, with respect to B and overall. What is
the half life of A in reaction ?
(I.I.T.)
Solution. From data (i) and (ii) it is obvious
that when the concentration of B is kept
constant (0.01 mol litre -1 ) and the
concentration of A is doubled (0.01 to 0.020
mol litre -1 ), the rate of reaction is also doubled
(0.005 to 0.010 mol litre -1 min -1 ). This shows
that the rate of reaction veries directly as the
first power of the concentration. Hence the
order of reaction with respect A is 1.
Similarly from data (i) and (iii) it is obvious
that when the concentration of A is kept
constant (0.01 mol litre -1 ) and the
concentratio of B doubled (0.01 to 0.02 mol
litre -1 ), the rate reaction remains constant
(0.005 mole litre -1 min -1 ). This show that the
order of reaction with respect to B is zero.
Now we know that the rate of reaction
A+B→products is given by
Rate r = K[A] 1 [B] 0
r=K[A]
∴ K= r
[A]
= 0.005
0.01
= 0.5 min −1
doubled, what will be the effect on the rate of
reaction ?
If
Solution From the reaction it is evident that
− dc ∝ [A] [B]2
dt
concentration of A is doubled the rate of
reaction will increase to 2 times. Similarly,
when [B] is doubled the rate of reaction will
increase to (2) 2 = 4 times.
. . . The overall increase in rate of reaction
= 2 x 4 =8 times
Illustration 8. (a) Determine the order of
reaction [A→ Products] from the following
data.
[A] in mol -1 Rate of reaction in mol l -1
min -1
(i) 0.01 0.005
(ii) 0.02 0.010
(iii) 0.03 0.015
(b) Determine the rate constant for the
reaction.
Solution (a) The data (ii) indicates that when
the concentration of A is doubled the rate of
reaction also doubles. Similarly, the data (iii)
indicates that when concentration of A is
made three times, the reaction rate also
becomes triple. Thus it is evident that the rate
of reaction is directly proportional to
concentration i.e.
− dx
dt α[A]
Hence the reaction is of first order.
(b) Rate = K [A]
or K= Rate
[A]
Here K = 0.005mol l−1 min −1
0.01 mol l −1 = 0.5 min −1
Illustration 9. The first order rate constant for
the decomposition of N 2
O 5
is 6.2 x 10 -4
second -1 . Calculate the half life period in
second for this decomposition
(14)

AISECT TUTORIALS : CHEMISTRY : SET-6
[M.L.N.R.]
Solution. We know that for a first order
reaction
0.693
t 1
/ 2
= = 1117.74 sec onds
6.2 x 10 −4
Illustration 10. In the reaction
A + 2B → 6C + 2 D
if the initial rate - − d[A]
at t =0 is 2.6 x 10 -2 m
dt
sec -1 , what will be the value of at t=0 ?
− d[B]
dt
Solution. From the reaction it is evident that
when a mole of A is reacting 2 moles of B
must react. Hence the decrease in the
concentration of B must be twice that of A
. . . − d[B]
dt
d[A]
= 2 ⎡ ⎤ ⎣ dt ⎦
=2 x 2.6 x 10 -2
= 5.2 x 10 -2 m sec -1
11.DETERMINATION OF ORDER OF
REACTION
The various important methods for the
determination of order of reaction are :
(a) Integration Method - In this method,
known amounts of reactants are mixed and
the progress of reactions is determined by
analysing the reaction mixture from time to
time. The data thus obtained are substituted
in the kinetic equations of first second and
third orders. Order of the reaction is then
known from that equation which gives a
constant value of rate constant k 1
, k 2
or k 3
.
Thus
2.303
k 1
= t
log a
(a−x) (for First order reaction)
1 x
K 2
= (for Second order reaction)
K3 =
⎡ t ⎣
⎤ a(a−x) ⎦
1 ⎡ x(2a−x) ⎤
2t ⎣ a 2 (a−x) 2 ⎦
(for Third order reaction)
(b) Graphical Method- If a straight line is
obtained by plotting log (a-x) against time or
dx/dt., it is a first order reaction. Similarly, if
a straight line is obtained by plotting (a-x) 2 or
(a-x) 3 against the reactants are at the same
initial concentraion.
(c) Half life Method - In general, the time for
50% change in concentration of the
reactants. known as time for half change is
inversely proportional to (n-1) power of the
initial concentraion. Thus
t 1 / 2 ∝ 1
a n−1
where n is the order of reaction. Hence t 1
/ 2
is
independent of initial concentration α (Half life
∝ to a 0 ) in case of First order reaction
inversely propotional to initial concentration a
(Half life ∝ to a -1 ) in case of second order
reaction and inversely proportioanl to square
of initial concentration in case of third order
reaction Half life ∝ a -2 ). Thus t 1
/ 2
a a 0 for first
order reaction t 1
/ 2
a 1/a for second order
reaction and t 1
/ 2
a 1 ∝/a 2 for the third order
reaction.
(d) Ostwald Isolation Method- This method
is based on the fact that if a reaction involves
nA molecules of A, nB molecules of B and nC
molecuels of C, the total order is n A
+ n B
+ n C
.
When all but one of the reactants in turn are
taken in excess so that their acive masses
remain constant through out the change the
concentration changes only of the isolated
reactant only. Thus when B and C are in
excess, the order of reaction will be nA, which
can easily be calculated. Similarly nB and nC
can be determined by taking the reactants A
and C and then A and B respectively in
excess. For example,
2CH 3
COOAg + HCOONa → 2 Ag + CO 2
+
CH 3
COOH + CH 3
COONa
(15)

AISECT TUTORIALS : CHEMISTRY : SET-6
In this reaction, when silver acetate is in
excess, the reaction is first order w.r.t.
sodium formate. When sodium formate is
taken in large excess the reaction is second
order w.r.t. silver acetate. Hence the total
order of the reaction is 1+2=3.
12.THEORY OF REACTION RATES
The theory of reaction rates (a) Explains why
the rates of various reactions may differ by
many powers ofter and (b) attempts to predict
the rates of various reactions from basic
ideas.
There are two important theories. The
collision theory of bimolecular reactions
and transition state or activated complex
theory. The latter is also known as
absolute reaction rate theory.
(a) Collision Theory - The main points of
this theory are:-
Fig. 2
(1) If two molecules are to react together
they must collide together.
(2) All collisions do not lead to chemical
reactions. Only those collisions give rise
to chemical reaction in which the
molecule acquires energy greater than
the activation energy. Thus only those
collisions result in product formation in
which the colliding molecules are
associated with a certain minimum
amount of energy. This minimum
energy which the molecule should
possess so that their mutual collisions
result in chemical reaction is called
threshold energy. The collisions which
result in the formation of product are
called effective collision,. Collision
among molecules possessing energy
less than the threshold energy are not
the effective collisions and so do not
result in the formation of products.
Thus colliding molecules must possess
cerlain minimum energy (Threshold
energy) to make the collision effective,
but most molecules, called normal
molecules have less energy than the
threshold energy. The additional energy
required by -the molecule to attain
threshold energy is called activation
energy, which is acquired by the
molecules as a result of interchange
(energies during the collisions. Hence,
Activation Energy = Threshold Energy-
Energy of Collidig Molecules,
The minimum excess energy that the normal
molecules must possess in order that the
collisions between them lead to chemical
reaction is called energy of activation. The
Threshold energy is always greater than
activation energy.
(b)Transition State Theory- it is based on
the idea that bond breaking and bond making
involved in a chemical reaction must occur
continuously or simultaneously. For example,
reaction between one hydrogen and one
(16)

AISECT TUTORIALS : CHEMISTRY : SET-6
iodine molecule to form 2 hydrogen iodide
molecules.
H 2
+I 2
→2HI
At some state in the process the H -H and I -
I bonds must be ruptured, while the H-I bonds
are being established. Thus if we represent a
partially ruptured or established bond by a
dotted line, we can write,
H-H H-H H H
I I I I → I + I
I I I I I I
Reactants Activated complex Products
Similarly reaction between CO and N0 2
can
be represented as,
≡
≡
O C+0-N → 0 C--- O---N → 0= C= 0+ N =O
II
II
0 0
Reactants Activated Complex Products
The dotted lines represent partial bonds,
The intermediate product with the partially
formed bond is called activated complex or
transition state, and the energy of activation is
the energy required to form the activated
complex or intermediate. (That is, the
difference in energy between the activated
complex and the reactant molecules). in the
reaction between CO and N0 2
the activated
complex has an energy about 32 k.cal greater
than that of reactants, CO and NO 2
and 88
k.cal greater than that of products. this
indicates that energy of activation of forward
and backword reaction is 32 kcals and 88
k.cals respectively. the difference between
these two quantities is the energy or entholpy
difference ( ∆H) between the products and
reactants. For an exothermic reaction ( ∆H <
0) and the energy of activation for the reverse
reaction is greater than that of forward
reaction. If the energy of activation of
products is greater than that of reactants ( ∆H
> 0) and the reaction is said to be
endothermic. Moreover, the molecularity of
the reaction is the number of molecules
which go into forming the activated
complex.The energy of the activated complex
will be higher than that of the reactants and
products. Hence, the reactants are not
converted directly into the products. There is
an energy barrier or activated complex
between the reactants and products.The
reactants must cross this energy barrier
before converting into products. The height of
the barrier determines the threshold energy.
The bonds in the activated complex shown by
dotted lines show that they are slightly longer
than the ordinary covalent bonds.The
activated complex may breakdown either into
reactants or products, which are therfore in
equilibrium with each other. In such a case
the energy of activation becomes the
enthalpy change in forming the activated
complex.
WHY ONLY A FRACTION OF
COLLISIONS LEADS TO REACTION?
Only small fraction of the collisions that
occurs actually leads to reaction
between the colliding species. This is
due to the fact that a molecule has no
definite boundary and there is fairly
diffuse electron cloud surrounding all the
nucleL When two molecules approach
closely to each other mutual repulsion of
their electron clouds takes place. As a
result of this repulsion, they slow down
and the consequent decrease in their KE
is accompanied by an increase in
potential energy. If the molecules were
(17)

AISECT TUTORIALS : CHEMISTRY : SET-6
already moving slowly before the
collision, they would stop, change
direction and move apart before any
significant interpenetration of their
electron clouds could take place. In
other words, slow moving molecules
simply bounce off each other and does
not produce any reaction. We therefore
conclude that only a collision between
molecules with sufficiently large speeds
can result in chemical reaction. In
addition, there is another factor that is
also taken into consideration in
determining whether or not a collision
between two molecules is effective at
leading to reaction. This factor is
orientation factor. In order for a reaction
to take place, specific bonds must be
broken and the colliding molecules must
be oriented w. r. t one another so that
new bonds may be formed Thus there
are two important reasons which are
responsible for the fact that only a very
Small fraction of collisions result in
reaction. These are
(1) The colliding molecules may not be
properly oriented to one another.
(2)The collisions not be sufficiently
energetic.
The minimum energy that must be
available for a collision to lead to a
reaction is known as activation energy
and if two molecules collide with less
energy than the activation energy, they
will recoil without undergoing chemical
change. The KE of the collision must be
transformed into potential energy.
Translational energy is converted into
rotational and vibrational energy, and as
atoms within a molecule vibrate with
large amplitude, bonds are expected to
be broken. The activation energy causes
the rupture of bonds between atoms in a
molecule or between an ion and its
solvation shel! if the reacting species are
in solution.
When far apart, hydrogen as well as iodine
molecules are quite stable and the potential
energy of the system E R
is minimum. When
these molecules approach each other, the
system becomes unstable and as a result
potential energy of the system increases and
attains a maximum value E A
when the
activated complex is formed. The potential
energy of the system again starts decreasing
when the activated complex breaks into two
molecules of HI. The potential energy goes
on decreasing till the two molecules of HI are
sufficiently apart and another minimum
potential energy (maximum stability) E p
is
obtained for the product molecules. The E A
-
E R
is called the energy barrier for the
forward reaction and is equal to the
energy of activation for the forward
reaction. Similarly, the energy of activation
for the backward reaction would be E A
-E p
.
For Forward Reaction,
H 2
+l 2
→ 2HI,
E p
-E R
= ∆E
For Backward Reaction, 2HI → H 2
+ I 2
E p
-E R
= −∆E
This indicates that forward reaction is
endothermic and backward reaction is
exothermic. Here it should also be noted that
energy of activation for a reaction is also the
difference between potential energy of the
activated complex and the potential energy of
the reactants.
(18)

AISECT TUTORIALS : CHEMISTRY : SET-6
The magnitude of the energy of activation A = Frequency factor constant,
10 0 rise of temperature. The quantitative
K=A
relation between K and T is given by
Arrhenius equation.
(ii) When Ea of a reaction is zero, the
reaction rate becomes independent of
K=Ae -Ea/RT temperature.
accounts for the rate of a chemical reaction.
Smaller the activation energy, greater will be
the rate of reaction and vice-versa
preexponential factor
E a
= Energy of activation
The term e -Ea/RT is called Boltzman factor.
13. Influence of Temperature on This factor represents the fraction of
Reaction Rate.
Increase in temperature increases the rate of
reaction. The change of rate with temperature
molecules having energy in excess of
activation energy. Arrhenius equation may
also be written as
is expressed by a change in the specific rate
E a
log K = log A -
constant, K. For every reaction, K increases
2.303 RT
with temperature. In several reactions (not in A plot of log K vs I/T gives a straight line
all), a 10 0 E
C rise in temperature approximately
a
whose slope is equal to
2.303 R
doubles or triples the reaction rate. The ratio
If rate constant for reaction at two different
of the specific rate constant of reaction at
temperatures is known, energy of activation
two temperatures separated by 10 0 C
can be calculated from the equation
(generally at 25 0 C and 35 0 C) is known as the
log K 2 E
temperature coefficient.
log K 1
=
a
2−T 1
2.303 R T 1 T 2
Temperature coefficient
For a reaction whose temperature
= K at(t+10 0 C)
= K coefficient is 2, if the reaction temperature is
35 0 C
K at t 0 C K raised from 25 0 C to 65 0 C, the rate increases
25 0 C
The temperature coefficient for most of the
reactions varies between 2 and 3. It means
by a factor of 2 4 (i.e. 16 times).
Note.
that the reaction rate becomes double for 10 0
rise in temperature. This is presumably
because the effective collisions double for 10 0
rise in temperature. It must be noted that
only reactions whose activation energy falls in
the 50-55 kj are found to double their rate for
(i) In the Arrhenius equation, when the
absolute temperature of the reaction
becomes infinity, the rate constant of
the reaction becomes equal to A, the
pre-exponential factor.
K = Ae −E a/RT = Ae
0
(19)

AISECT TUTORIALS : CHEMISTRY : SET-6
14. SYNOPSIS & EXPLANATORY
NOTES
Chemical Kinetics predicts the rate and
mechanism of chemical reactions.
The rate of reaction is the rate at which
the concentration of the reactants
decreases or the concentration of
products increases. The unit of rate of
reaction is moles litre -1 sec -1
An equation which relates the rates of a
reaction to the concentrations of the
reactants is called the rate law or rate
equation. Order of reaction is the
number of molecules whose
concentration alters as a result of
chemical change. Rate constant, or
velocity constant is the rate of reaction
when the concentrations of reactants are
equal to unity. Zero order reactions are
those in which the rate of reaction is
independent of the concentration of the
reactants. The unit of zero order
reaction is moles litre -1 sec -1 (same as
that of rate of reaction). The various
factors which are known to affect the
speed of a reaction are (a)
Temperamre (b) Concentration of
reactants (c) The surface area of the
reactants (d) Electromagnetic
radiation, and (e) The pressence of a
catalyst or enzyme. In addition,
pressure can affect the gaseous
reactions mainly. Radiation other than
the electromagnetic type (e.g. visible
light, X-rays, gamma rays, infra red, ultra
violet, etc) e.g, beam of protons.
neutrons, alpha particles, beta particle
etc can also affect reaction rates.
Reactions are also affected by the
nature of the solvent and by the
presence of ions if the reaction is ionic.
Raising the temperature raises the
speed of reaction.
Rate of reaction reciprocal of time ccl
concentration of reactants.
The concentration of a solution is
essentially the number of particles per
unit volume, and for gaseous reactions
an increase in pressure amounts to an
increase in concentration, since the
gaseous molecules are brought nearer
together, and will simply give rise to a
faster reaction.
If a substance is in finely divided form its
surace area is much increased. Hence
rate of reaction increases with
subdivision. A catalyst is a substance
which alters the rate of chemical reaction
without itself being chemically changed.
One explanation of why the above
factors affect the rate of a reaction is the
collision hypothesis, which states that
the rate of a reaction depends upon the
number of molecular collisions between
reactants, not every collision leading to
reaction. An increase in temperature
causes the molecules to travel faster,
and hence increases the probability of
them colliding. An increase in
concentration or pressure increases the
number of molecules per unit volume,
and hence increases the probability of
them colliding. If the solid reactants have
a large surface area, more atoms are
allowed to react. Hetrogeneous
catalysts allow molecules to reside on
(20)

AISECT TUTORIALS : CHEMISTRY : SET-6
their surface in such an exposed
condition that majority of collisions bring
about reaction. Homogeneous
catalysts allow collisions to take place in
such a manner that two or more simple
reactions are substituted for one
complicated reaction, which requires a
lot of energy to occur.
The only satisfactory way of finding out
how a reaction rate depends upon
concentration of its reactants is through
carefully designed experiments.
Reactions proceed faster at higher
temperatures because only reactant
molecules possessing enough energy to
allow making and breaking of bonds to
occur can react, on meeting, and this
fraction of molecules with enough
energy usually increases greatly with
increase in temperature.
The rate or velocity constant of a
reaction is the change in concentration
per unit time of reactant or product in a
reaction in which all the reactants are at
unit concentration.
By convention, the reaction products
are placed in the numerator and the
reactants in the denominator.
Concenfrations used in such
expressions must be those in the
equilibrium mixture.
Since a catalyst alters the rate of a
chemical reaction, it also alters the rate
constant of the reaction. A catalyst,
however, has no effect on the
composition of the equilibrium mixture in
a reversible reaction and hence does not
alter the value of equilibrium constant. ft
does speed up the forward and
backward reaction to equal extents.
Hence equilibrium is established more
rapidly in presence of a catalyst. if the
equilibrium constant was affected, the
law of conservation of energy will not be
satisfied.
X →
← Y, heat evolved
Addition of catalyst would shift the
equilibrium from left to right and an
equilibrium mixture of X and Y could
therefore be made to evolve heat, simply
by adding the catalyst.
Pressure has very little effect on
solids and liquids and consequently
affects only very slightly any reaction
involving solids and or liquids. For
reactions involving gases, an increase in
pressure increases the concentration by
bringing the moecules closer together
and hence increases the rate of reaction.
The rate constant is not, however,
altered by pressure. Change in pressure
does not cause any change in the
equilibrium constant of a reaction which
is reversible, but it does alter the
composition of the equilibrium mixture of
any reaction in which a change of
volume occurs. For reactions that takes
piace with a decrease in volume (e. g. N 2
+ 3H 2
= 2NH 3
or 2SO 2
+ O 2
⇌ 2SO 3
)
an increase in pressure causes a shift in
the equilibrium from left to right, while for
reactions which occur with increase in v
olume (e. g. PCl 5
= PCI 3
+ C1 2
or N 2
0 4
⇌ 2NO 2
increase of pressure shifts the
equilibrium from right to left. For
reactions in which there is no change in
(21)

AISECT TUTORIALS : CHEMISTRY : SET-6
volume (H 2
+l 2
=2HI), pressure has no
effect on the composition of equilibrium
mixture.
The rate of a reaction is increased by
increasing the temperature and the
velocity constant is only constant at
constant temperature. Increase of
temperature will increase molecular
motion and increases the rate of
intermolecular collisions. For most
reactions, the velocity constant is almost
doubled for a 10 o C rise in temperature.
This means an increase of about 10%
for every I o C rise in temperature. It has
been found that the total molecular
collisions increase by only about 2% for
every 1 0 C rise in temperature.
Moreover, the total number of collisions
at any temperature is greater than the
number of molecules which actually
undergo reaction. This means that a
certain portion of the total molecules
actually collide and react. This proportion
increases more rapidly with increase in
temperature than does the number of
collisions.
It is uncommon for reactions to take
place in stages involving trimolecular
steps. No reaction is known having
molecularity greater than three. Majority
of reactions involve unimolecular or
bimolecular ste
ps only. The molecularity of a reaction
must always be a whole number.
Molecularity gives no information about
the rate of a reaction. Experiments show
that reaction rates nearly always depend
in some manner on the concentration of
reactants.
The order of reaction usually assumes
values of 1 or 2, occasionally 0 to 3, and
sometimes an obviour fraction such as
1/4, 1/2, 2/3 or 3/4. The order of a
reaction need not be whole number. It is
also not necessarily related, numerically,
to the chemical equation written for the
reaction. For each basic step in the
reaction, the order of reaction and
molecularity will generally be the same,
but for complex reactions they will
normally differ. The overall kinetics of
the reaction depends upon what is
known as rate determining step. When
a reaction takes place in stages, the rate
is determined by the slowest stage,
knows as rate determining step.
First order reaction is one in which the
rate is directly proportional to the
concentration of reacting substance or
rate is determined by the variation of one
concentration term. The unit of first order
reaction is sec -1 or time -1
A second order reaction is one in which
the rate is determined by the variation in
two concentration terms. The unit of
second order reaction is litre.mole -1 .
sec -1.
A Second order reaction gives first order
result when one of the reactants is
present in excess.
A third order reaction is one in which the
rate is determined by the variation of
three concentration terms. The unit of
third order reaction is moles -2 .litre -2 . sec -1
(22)

AISECT TUTORIALS : CHEMISTRY : SET-6
The half life for a first order reaction is
independent of the initial concentration.
For a second order reaction, it is
inversely proportional to the initial
concentration, and for a third order
reaction it is inversely proportional to the
square of the initial conc~ntration. In
general, the half life times t 1
and t 2
for a
reaction with initial concentrations, c 1
and c 2
are related as
t 1
t 2
= ⎛ ⎝ c 2
c 1
⎞ ⎠
n−1
where n is the order of reaction.
For a fast reaction, for which k is large,
the half life (T 1
/ 2
) will be small and for a
slow reaction, for which k is small, the
half life will be comparatively large
because half life (T 1
/ 2
) is inversely
proportional to the rate constant k,
according to the equation T 1
/ 2
= 0.693/k.
The minimum energy which the
molecules should possess so that their
collisions result in chemical reaction is
called threshold energy. The collisions
which result in the formation of products
are called effective collisions.
Collisions among molecules possessing
energy less than the threshold energy
are not the effective collisions and so do
not result in the formation of products.
The additional energy required by the
molecules to attain threshold energy is
called activation energy, which is
equal to Threshold energy-Energy of
colliding molecules. Thus the
minimum excess energy that the normal
molecules must possess inorder that the
collision between them lead to chemical
reaction is called energy of activation.
The threshold energy is always
greater than energy of activation.
In order to react, a molecule, or a pair of
molecules in collision, must posses a
certain minimum energy, called the
energy of activation. For a reaction
which proceeds only very slowly at room
temperature this activation energy is well
above the average K.E. of the molecule.
Before a molecule can react it must
acquire energy either by a series of lucky
collisions or by absorption of energy
(infra red, ultraviolet or other radiation).
For an activation energy E, the number
of molecules with energy greater than E
will be n. Hence rate of reaction and
velocity constant of a reaction will be
proportional to n. Thus rate of reaction
K= A.e -E(RT), where A is constant known
as Arrhenlous factor or frequency
factor. The equation K = A.e -E(RT) is
known as Arrhenius equation. Increase
in temperature increases the number of
activated molecules to a much greater
extent than the number having average
K.E.
The less the energy of activation for a
reaction, the more easily that reaction
will take place.
A very few reactions of an elementary
nature show a rate constant decreasing
with rise in temperature, E approaching
to be negative, and lying between 0 to-4.
Reactions generally proceed more
readily when the products have a lower
energy content than the reactants, the
difference being the free energy of
reaction, ∆H.
(23)

AISECT TUTORIALS : CHEMISTRY : SET-6
In order for a reaction to proceed,
energy has to be supplied to the
reactants in order to carry them over
activated complex, this energy being
required, for the most part, to stretch
and rupture any bonds as may be
necessary in the reacting molecules.
This process proceeds readily for
activated molecules. The amount of
work necessary to get the reactants up
to the top of the activated complex is
called free energy of activation, ∆G
which we are calling simply the energy of
activation.
Highly endothermic reactions are
unlikely to occur readily at low
temperature. The energy of activation
has to be greater than the heat of
reaction so that the activation energy
for any endothermic reaction with a
high heat of reaction must be high.
The theory of reaction rates (a)
Explains why the rates of various
reactions may differ by many powers
of 10, and (b) Attempts to predict the
rates of reactions from basic ideas.
Collision theory is based on the
principle that if two molecules are to
react together they must collide together.
Then, using the activation energy, it is
postulated that not only every collision is
fruitful as regards producing chemical
reaction, only collisions in which the
molecule acquires energy greater than
the activation energy doing so. Hence
collision theory states that the rate of
a reaction is equal to the number of
collisions between activated species
in one second.
The transition state theory or activated
complex theory is based on the idea that
bond breaking and bond making
involves a chemical reaction which must
occur continuously and simultaneously
rather than something which happens
instantaneously. For example, in the
reaction between one hydrogen and one
iodine molecule gives rise to two
hydrogen iodide molecules. At some
stage of the process, the H-H and I-I
bonds must be vuptured while the H-I
bonds are being established. The
intermediate product with the partially
formed bonds is called activated
complex or transition state, and the
energy of activation is the energy
required to form this intermediate. The
molecularity of the reaction is the
number of molecules which go into the
forming of activated complex.
(24)

AISECT TUTORIALS : CHEMISTRY : SET-6
Objective Questions
CHEMICAL KINETICS
1. The rate of a chemical reaction
(a)
(b)
(c)
(d)
[MP PMT; CPMT]
Increases as the reaction proceeds
Decreases as the reaction proceeds
May increase or decrease during the
reaction.
Remains constant as the reaction
proceeds
2. The specific rate constant of a first
order reaction depends on the
(a)
(b)
(c)
(d)
[IIT; Delhi PMT; MP PAT;
Bihar MEE; Karnataka CET]
Concentration of the reactants
Concentration of the products
Time of reaction
Temperature of reaction
3. According to the collision theory of
chemical reactions
(a)
(b)
(c)
(d)
A chemical reaction occurs with every
molecular collision
Rate is directly proportional to the
number of collisions per seconds
Reactions in the gas phase are always
of zero order
Reactions rates are of the order of
molecular speeds.
4. If the concentration is expressed in
moles per litre the unit of the rate
constant for a first order reaction is
[MLNR; MP PET; Bihar MEE;
CPMT; MP PMT]
(25)
(a) mole litre -1 sec -1
(b) mole litre -1
(c) sec -1
(d) mole -1 litre -1 sec -1
5. The unit of rate constant of second
order reaction is usually expressed as
(a) mole litre sec -1
(b) mole -1 litre -1 sec -1
(c) mole litre -1 sec -1
(d) mole -1 litre sec -1
[NCERT; MLNR; MP PMT]
6. A zero order reaction is one whose rate
is independent of
(a)
(b)
(c)
(d)
Temperature of the reaction
The concentration of the rectants
The concentration of the products
The material of the vessel in which the
reaction is carried out.
7. Which of the following rate laws has an
overall order of 0.5 for reaction involving
substances x, y and z
(a) Rate = K(C x
) (C y
) (C z
)
(b) Rate = K(C x
) 0.5 (C y
) 0.5 (C z
) 0.5
(c) Rate= K(C x
) 1.5 (C y
) -1 (C z
) 0
(d) Rate= K(C x
) (C z
) n / (C y
) 2
[AIMS]
8. For the reaction A+2B → C the rate of
reaction at a given instant can be
represented by
(a)
+ d[A]
= − 1 d[B]
dt 2 dt
= + d[C]
dt

AISECT TUTORIALS : CHEMISTRY : SET-6
(b)
(c)
(d)
− d[A]
dt
= + 1 2
− d[A]
dt
= − 1 2
+ d[A]
dt
= + 1 2
d[B]
dt
d[B]
dt
d[B]
dt
= + d[C]
dt
= + d[C]
dt
= + d[C]
dt
9. The correct order indicated against the
K
rate of reaction A +B → is
(a)
d[C]
dt
= K [A]
(b) − d[C]
= K[A] [B]
dt
(c) − d[A]
dt
= K[A] [B]
(c) − d[A]
dt
= K[A]
[B.H.U.]
10. In a reaction 2A + B → A 2
B the reactant
A will disappear at
(a)
(b)
(c)
(d)
[M.P.P.E.T.]
half the rate at that B will decrease
the same rate at that B will decrease
the same rate at the A 2
B will form
twice the rate at that B will decrease
11. If the concentration of the reactants is
increased the rate of reaction
(a)
(b)
(c)
(d)
remains unaffected
increases
decreases
may increases or decrease
[M.P. BOARD]
12. The specific rate constant of a first
order reaction depends on the
(a)
(b)
(c)
(d)
Concentration of the reactant
Concentration of the product
Time
Temperature
[I.I.T., Modified D.P.M.T.]
13. The rate constant of a reaction depends
on
(a) Temperature (b) Mass
(c) Weight (d) Time
14. Which of the following statements
regarding the molecularity of a reactions
is wrong ?
(a)
(b)
(c)
(d)
It is the number of molecules of the
reactants taking part in a single step
chemical reaction.
It is calculated from the reaction
mechanism
It may be either a whole number of
fractional
It depends on the rate determining step
in the reaction
[C.B.S.E.]
15. Which of the following stands true for
order of a reaction ?
(a)
(b)
(c)
(d)
It is equal to the sum of exponents of
the molar concentrations of the
reactants in the rate equation.
It is always a whole number
It is never zero
It is a theoretical concept which
depends on the rate determining step
reaction in the reaction mechanism.
16. Which one of the following statements
about the order of a reaction is true ?
(a)
(b)
(c)
The order of a reaction can only be
determined by experiment.
The order of a reaction increases with
increase in temperature
The order of a reaction can be
determined from the balanced equation
(26)

AISECT TUTORIALS : CHEMISTRY : SET-6
(d)
A second order reaction is also
bimolecular
[K.C.E.T.]
17. A zero order reaction is one whose rate
is independent of
(a)
(b)
(c)
(d)
Temperature of the reaction
[N.C.E.R.T.]
The concentration of the reactants
The concentration of the products
The material of the vassel in which the
reaction is carried out.
18. For a zero order reaction
(a)
(b)
(c)
(d)
The reaction rate is double when the
initial concentration is doubled
The time for half change is half the time
taken for completion of the reaction.
The time for half change is independent
of the initial concentration.
The time for completion of the reaction
is independent of the initital
concentration.
19. The one which is unimolecular reaction
is
(a) 2HI → H 2
+ I 2
(b) N 2
O 5
→ N 2
O 4
+ 1 2 O 2
(c)
H 2
+ Cl 2
→2HCl
(d) Pcl 3
+ Cl 2
→ PCl 5
[M.P.P.A.T.]
20. The hydrolysis of ethyl acetate
CH 3
COOEt + H 2
O
a reaction of
H +
→
CH 3
COOH+ EtOH is
[M.P. PET]
(a) First order (b) Second order
(c) Third order (d) Zero order
21. The order of a reaction of a radioactive
decay is
(a) Zero (b) Two
(c) Three (d) one
[Orissa M.B.B.S.]
22. Diazonium salt decomposes as
(a)
(b)
(c)
(d)
+
C 6
H 5
N + Cl - 2 → C 6
H 5
Cl + N 2
At O 0 C, the evolution of N 2
becomes
two times faster when the initial
concentrations of the salt is doubled.
Therefore it is
a first order reaction
a second order reaction
independent of the initial concentraiton
a zero order reaction
[M.L.N.R.]
23. Hydrolysis of ester in alkaline medium is
(a)
(b)
(c)
(d)
First order reaction with molecularity
one.
second order reaction with molecularity
two.
first order reaction with molecularity
two.
second order reaction with molecularity
one.
[A.F.M.C.]
24. For a reaction; mA + nB → products the
rate is given by
⎯⎯
−∆ [A]
∆ t
= K[A] m [B] n →
(where m and n are constant numbers
for the reaction, and t is the time). The
overall order of the reaction will be
equal to
(27)

(a) m (b) n
(c) m-n (d) m +n
25. For the reaction: H 2
(g)+Br 2
(g)→2
Hbr(g), the experimental data suggests
Rate = K [H 2
] [Br 2
] 1/2
(a)
The molecularity and oirder of reaction
for the reaciton is
2 and 2 respectively
(b) 2 and 1 1 respectively
2
(c) 1 1 and 2 respectively
2
(d) 1 1 and 1 1 respectively
2 2
26. The rate law for the reaction
RCl + NaOH → ROH + NaCl
[M.P.P.E.T.]
is given by Rate =K[RCl]. The rate of
this reaction.
(a) is doubled by doubling the
concentration of NaOH
(b)
is halved by reducing the concentration
of RCl by one half
(c) is increased by increasing the
temperature of the reaction
(d)
in unaffected by change in temperature
which is correct ?
(a) A and B (b) B and C
(c) C and D (d) B and D
27. For the following reaction scheme
(homogeneous), the rate constant has
units.
A + B
K
→
C
(a) sec -1 (b) sec -1 mole
(c) Sec -1 mole -1 (d) sec.
AISECT TUTORIALS : CHEMISTRY : SET-6
[KCET]
(28)
28. In a catalytic conversion of N 2
to NH 3
by
Haber's process, the rate of reaction
expressed as change in the
concentration of ammonia per time is
40 x 10 -3 mole -1 s -1 . If there are no side
reactions, the rate of the reaction as
expressed in terms of hydrogen is
(a) 60 x 10 -3 mol l -1 s -1
(b) 20 x10 -3 mol l -1 s -1
(c) 1200 mol 1 -1 s -1
(d) 10.3 x 10 -3 mol l -1 s -1
29. In the reaction of A + 2B → C +2D the
initial rate, If at t=0 was found to be
2.6 x 10 -2 M se c-1 . What is the value of
If at t =0 in M s -1 .
(a) 2.6 x 10 -2 (b) 5.2 x 10 -2
(c) 1.0 x 10 -1 (d) 6.5x10 -3
30. One litre of 2 M acetic acid is mixed
with one litre of 3 M ethyl alcohol to
form ester
CH 3
COOH+C 2
H 5
OH →CH 3
COOC 2
H 5
+H 2
O
The decrease in the initial rate, if each
solution is diluted by an equal volume of
water would be
(a) 0.25 times (b) 0.5 times
(c) 2 times (d) 4 times
31. What will be the order of the reaction if
doubling of the concentration of a
reactant increases the rate by a factor
of 4 and trebling the concentration of
the reactant by a factor of 9 ?
(A.I.I.M.S.)
(a) First order (b) Zero order
(c) Second order (d) Third order
32. For a reaction : 2A + B → Products the
active mass of B is kept constant and

AISECT TUTORIALS : CHEMISTRY : SET-6
(a)
(b)
(c)
(d)
that of A is doubled. The rate of
reaction will then
Increase two times
Increase four times
decrease two times
decrease four times
[M.P.P.E.T.]
33. Which of the following statement is not
correct for the reaction ?
(a)
(b)
(c)
(d)
4 A + B ⇌ 2C + 2D
[A.I.I.M.S.]
The rate of disapperance of B is twice
the rate of appearance of C
The rate of disapperance of B is one
fourth the rate of disapperance of A
The rate of formation of D is one half
the rate of consumption of A.
The rate of formation of C and D are
equal.
34. If the concentration is expressed in
moles per litre, the unit of the rate
constant for a first order reaction is
(a) Mole litre -1 sec -1
(b) Mole litre -1
(c) Sec -1
(d) Mole -1
[M.L.N.R.]
35. The second order rate constant is
usually expressed as
(a) Moles litre sec -1
(b)
Mole -1 litre
(c) Mole litre -1 sec -1
(d) Mole -1 litre sec -1
36. The rate of reaction
Cl 3
.CHO + NO → CHCl 3
+ NO + CO is
given equation,
Rate = K[Cl 3
.CCHO][NO].
If concentration is expressed in
moles/litre, the units of K are
(a) Litre -2 mole -2 sec -1
(b) mole litre -1 sec -1
(c) litre mole -1 sec -1
(d) sec -1 (MP PET)
37. The unit of rate constant for a zero
order reaction is
(a) Litre sec -1
(b) Litre mole -1 sec -1
(c) Mole litre -1 sec -1
(d) Mole sec -1
(NCERT)
38. A reaction involving two different
reactants
(a)
(b)
(c)
(d)
can never be a first order reaction
can never be a second order reaction
can never be a unimolecular reaction
can never be a bimolecular reaction
(KCET)
39. The specific rate for a reaction is 1.0 X
10 -4 mol lit -1 min -1 . The order of the
reaction is
(a) zero (b) one
(c) two (d) three
40. The half life of a first order reaction is
69.35. The value of the rate constant of
the reaction is
(a) 1.0 s -1 (b) 0.1 s -1
(CBSE)
(29)

AISECT TUTORIALS : CHEMISTRY : SET-6
(c) 0.01 s -1 (d) 0.001 s -1
41. A first order reaction has specific rate
constant of 2 min -1 . The half life of the
reaction will be
(a) 1.653 min (b) 0.347 min
(c) 2 min `(d) 0.0347 min
(Kurukhetra)
42. If 75% of first order reaction is
completed in 32 min, then 50% of the
reaction would be completed in
(a) 16 min (b) 24 min
(c) 10 min (d) 20 min
(EAMCET)
43. The minimum energy necessary to
permit a reaction is
(NCERT)
(a) Internal energy (b) threshold energy
(c) activation energy (d) free energy
44. The minimum energy required for the
reacting molecules to undergo reaction
is
(a) potential energy (b) kinetic energy
(c) thermal energy (d) activation energy
(KCET, EAMCET, MP PMT)
45. Energy of activation of an exothermic
reaction is
(a) zero (b) negative
(c) positive (d) can't be predicted
46. For an endothermic reaction, where ∆H
represents the enthalpy of the reaction
in kJ/mol, the minimum value for the
energy of activation will be
(a) less than ∆H (b) Zero
(I.I.T.)
(c) more than ∆H (d) equal to ∆H
47. If the reaction rate at given temperature
becomes slower, then
(a)
(b)
(c)
(d)
(MP PMT)
the free energy of activation is higher
the free energy of activation is lower
the entropy changes
the initial concentration of the reactants
remains constant
48. The following data are for the
decomposition of ammonium nitrate in
aqueous solution
Volume of N 2
in cc Time(minutes)
6.25 10
9.50 15
11.42 20
13.65 25
35.05 Finally
The order of the reaction is
(a) Zero (b) One
(c) Two (d) Three
[NCERT]
49. The rate law for the reaction
RCl + NaOH(aq) →ROH + NaCl is
given by Rate = K 1
[RCl]. The rate of the
reaction will be
(a)
(b)
[IIT]
Doubled on doubling the concentration
of sodium hydroxide
Halved on reducing the concentration of
alkyl halide to one half
(c) Decreased on increasing the
temperature of the reaction
(30)

AISECT TUTORIALS : CHEMISTRY : SET-6
(d) Unaffected by increasing the
temperature of the reaction
50. Which of the following is a first order
reaction
(a) NH 4
NO 2
→N 2
+2H 2
O
(b) 2HI→H 2
+I 2
(c) 2NO 2
→2NO+O 2
(d) 2NO+O 2
→2NO 2
[MP PMT]
51. The temperature coefficient of most of
the reactions lies between
(a) 1 and 3 (b) 2 and 3
(c) 1 and 4 (d) 2 and 4
[MP PET]
52. The inversion of cane sugar is
reperesented by
C 12
H 22
O 11
+ H 2
O →C 6
H 12
O 6
+ C 6
H 12
O 6
(a)
It is a reaction of
Second order
(b) Unimolecular
(c)
Pseudo unimolecular
(d) None of the three
[AFMC; MP PMT]
53. If doubling the concentration of a
reactant `A' increases the rate 4 times
and tripling the concentration of `A'
increases the rate 9 times, the rate is
proportional to
(a)
(b)
(c)
(d)
Concentration of `A'
Square of concentration of `A'
[AIIMS]
Under root of the concentration of `A'
Cube of concentration of `A'
54. The influence of temperature on the
rate of reaction can be found out by
(a)
(b)
(c)
(d)
Clapeyron-Claussius equation
Gibbs-Helmholtz equation
Arrihenius equation
Vander Waal's equation
55. Which one of the following formula
represents a first order reaction
(a)
K = x t
(b) K = 1 1
− 1 2t (a−x) 2 a 2
2.303 a
(c) K = t log 10 (a−x)
(d) K = 1 x
t a(a−x)
[MP PMT]
56. According to the collision theory of
reaction rates, rate of reaction
increases with temperature due to
(a)
Greater number of collisions
(b) Greater velocity of the reacting
molecules
(c)
(d)
Greater number of molecules have
activation energy
None of the above
57. The first order rate constant for the
decomposition of N 2
O 5
is 6.2 x 10 -4
sec -1 . The half life period for this
decomposition in seconds is
(a) 1117.7 (b) 111.7
(c) 223.4 (d) 160.9
[MLNR; MP PET]
58. The velocity of a chemical reaction with
the progress of reaction
(a)
(b)
Increases
Decreases
(31)

AISECT TUTORIALS : CHEMISTRY : SET-6
(c)
(d)
First increases and then decreases
Remain constant
59. The reaction rate at a given
temperature becomes slower, then
(a)
(b)
(c)
(d)
[MP PMT]
The free energy of activation is higher
The free energy of activation is lower
The entropy changes
The initial concentration of the reactant
remain constant
60. The rate of chemical reaction at
constant temperature is proportional to
(a)
(b)
(c)
(d)
The amount of products formed
The product of masses of the reactants
The product of the molar concentration
of the reactants
The mean free path of the reactants
61. The concentration of a reactant
decreases from 0.2 M to 0.1 M in 10
minutes. The rate of the reaction is
(a) 0.01 M (b) 10 -2
(c) 0.01 mol dm -3 min -1 (d) 1 mol dm -3 min -1
62. A first order reaction which is 30%
complete in 30 minutes has a half-life
period of
(a) 24.2 min (b) 58.2 min
(c) 102.2 min (d) 120.2 min
63. For the reaction
[AIIMS]
H
CH 3
COOCH 3
+ H 2
O →
+
CH 3
COOH +
CH 3
OH
The progress of the process of reaction
is followed by
(a)
(b)
(c)
(d)
Finding the amount of methanol formed
at different intervals
Finding the amount of acetic acid
formed at different intervals
Using a voltmeter
Using a polarimeter
→
←
64. The reaction 2N 2
O 5
2NO 2
+ O 2
(a)
(b)
(c)
(d)
follows first order kinetics. Hence, the
molecularity of the reaction is
unimolecular
Pseudo-unimolecular
Bimolecular
None of the above
65. A rise in temperature increases the
velocity of a reaction. It is because it
results in
(a)
(b)
(c)
(d)
An increased number of molecular
collisions
An increased momentum of colliding
molecules
An increase in the activation energy
A decrease in the activation energy
66. If E f
and E r
are the activation energies
of forword and reverse reactions and
the reaction is known to be exothermic,
then
(a)
(b)
(c)
(d)
E f
> E r
E f
< E r
E f
= E r
No relation can be given between E f
and E r
as data are not sufficient
67. A large increase in the rate of a reaction
for a rise in temperature is due to
[EAMCET; MP PET]
(32)

AISECT TUTORIALS : CHEMISTRY : SET-6
(a)
The decrease in the number of
collisions
(d)
The energy gained by the moleucles on
colliding with another molecule
(b) The increase in the number of
activated molecules
(c)
(d)
The shortening of the mean free path
The lowering of the activation energy
68. Consider the following energy profile for
the reaction. X + Y = R + S. Which fo
the following deductions about the
reaction is not correct
70. The velocity of the chemical reaction
doubles every 10 o C rise of temperature.
If the temperature is raised by 50 o C, the
velocity of the reaction increases to
about
(a) 30 times (b) 16 times
(c) 20 times (d) 50 times
71. By ``the overall order of a reaction', we
mean
(a)
(b)
(c)
(d)
The number of concentration terms in
the equation for the reaction
The sum of powers to which the
concentration terms are raised in the
velocity equation
The least number of moleucles of the
reactants needed for the reaction
The number of reactants which take
part in the reaction
(a)
(b)
(c)
(d)
Fig. 3
The energy of activation for the
backward reaction is 80 kJ
The forward reaction is endothermic
∆H for the forward reaction is 20 kJ
The energy of activation for the forward
reaction is 60 kJ
69. According to Arrhenius theory, the
activation energy is
(a)
(b)
(c)
The energy it should possess so that it
can enter into an effective collision
The energy which the moleucle should
possess in order to undergo reaction
The energy it has to acquire further so
that it can enter into a effective collison
72. Velocity constant K of a reaction is
affected by
(a)
(b)
(c)
(d)
Change in the concentration of the
reactant
Change of temperature
Change in the concentration of the
product
None of the above
73. Catalyst decomposition of hydrogen
peroxide is a...... order reaction
(a) First (b)Second
(c) Third (d) Zero
74. An increase in temperature by 10 o C,
generally increases the rate of a
reaction by
(33)

(a) 2 times (b) 10 times
(c) 9 times (d) 100 times
75. The velocity constants of a reaction is
K. Which of the following statements is
not true regarding K
(a)
(b)
(c)
(d)
K is a constant for a reaction at a given
temperature
The value of K changes when the
temperature changes
K is the velocity of the reaction at unit
concentrations of the reactant
K is a constant for all reactions
76. The energy of activation is
(a) The energy associated with the
activated molecules
(b)
(c)
Threshold energy- energy of normal
molecules
Threshold energy + energy of normal
molecules
(d) Energy of products - energy of
reactants
77. The half life of a first order reaction is
(a)
Independent of the initial concentration
of the reactant
(b) Directly proportional to the initial
concentration of the reactants
(c)
(d)
Inversely proportional to the initial
concentration of the reactant
Directly proportional to the square of the
intial concentration of the reactant
78. Which one of the following does not
reperesent Arrhenius equation
(a)
k = Ae -E/RT
E
(b) log e
k = log e
A-
RT
E
(c) log 10
k = log 10
A-
2.303RT
AISECT TUTORIALS : CHEMISTRY : SET-6
(34)
(d)
k = AE - RT
79. The decomposition of N 2
O 5
is a first
order reaction represented by N 2
O 5
→
1
N 2
O 4
+ O 2
. After 15 minutes the
2
volume of O 2
produced is 9 ml and at
the end of the reaction 35 ml. The rate
constant is equal to
(a)
(c)
1
15 In35 44
1
15 In44 35
1
(b)
15 In44 26
1
(d)
15 In35 26
[MP PET]
80. On increasing the temperature, the rate
of the reaction increases because of
(a)
(b)
(c)
(d)
[MP PMT]
Decrease in the number of collision
Decrease in the energy of activation
Decrease in the number of activated
molecules
Increase in the number of effective
collisions
81. The temperature coefficient for reaction
in which food deteriorates is 2. Then
food deteriorates .... times as rapidly at
25 o C as it does at 5 o C
(a) Two (b) Four
(c) Six (d) Twenty
82. The rate of a reaction is doubled for
every 10 o rise in temperature. The
increase in reaction rate as a result of
temperature rise from 10 o to 100 o is
(a) 112 (b) 512
(c) 400 (d) 614
[Karnatak CET]
83. The half life period of a first order
reaction

AISECT TUTORIALS : CHEMISTRY : SET-6
(a)
(c)
0.693
t
2.303
t
0.693
(b)
K
0.303
(d)
K 1
84. Energy of activation of a reactant is
reduced by
(a)
(b)
(c)
(d)
Increased temperature
Reduced temperature
Reduced pressure
Increased pressure
85. A catalyst increases the rate of a
chemical reaction by
(a)
(b)
(c)
(d)
Increasing the activation energy
[MLNR; CPMT]
Decreasing the activation energy
Reacting with reactants
Reacting with products
86. Velocity constant of a reaction at 290 K
was found to be 3.2 x 10 -3 . At 310 K it
will be about
(a) 1.28 x 10 -2 (b) 9.6 x 10 -3
(c) 6.4 x 10 -3 (d) 3.2 x 10 -4
[Karnataka CET]
87. Decay constant of a reaction is 1.1 x
10 -9 / sec, then the half life of the
reaction is
(a) 1.2 x 10 8
(b) 6.3 x 10 8
(c) 3.3 x 10 8
(d) 2.1 x 10 8
88. If the half life period of a reaction is
inversely proportional to the initial
concentration, the order of the reaction
is
(a) Zero (b) One
(c) Two (d) Three
89. Which one of the following statements
is wrong
(a)
(b)
(c)
(d)
Molecularity of a reaction is always a
whole number
Order and molecularity of a reaction
need not be same
Order of a reaction may be zero
Order of a reaction depends upon the
mechanism of the reaction
90. Which of the following statements is
not true according to collision theory of
reaction rates
(a)
(b)
(c)
(d)
Collision of molecules is a precondition
for any reaction to occur
All collisions result in the formation of
the products
Only activated collisions result in the
formation of the products
Moleucles which have acquired the
energy of activation can collide
effectively
91. The temperature coefficient of a
reaction is
(a)
(b)
Specific reaction rate at 25 o C
Rate of the reaction at 100 o C
(c) Ratio of the rate constants at
temperatures 35 0 C and 25 o C
(d)
Ratio of the rate constants at two
temperatures diffgering by 1 o C
92. The velocity constant of first order
reaction is expressed in the units
(a)
(b)
(c)
(d)
Concentration per unit time
Time per unit concentration
Per unit time
Unit time per unit concentration
93. For a zero order reaction
(35)

AISECT TUTORIALS : CHEMISTRY : SET-6
(a)
(b)
(c)
(d)
The concentration of the reactant does
not change during the reaction
The concentration change only when
the temperature changes
The rate remains constant throughout
The rate of the reaction is proportional
to the concentration
94. If `a' is the initial concentration and `n' is
the order of the reaction and the half life
period is `T,then
(a)
T ∝
n-1
∝
(b) T a n
(c) T ∝ 1 (d) T ∝ 1
a n a n−1
95. Half life period of a first order reaction is
138.6 minutes. The velocity constant of
the reaction is
(a) 0.05 min -1 (b) 0.00005 min -1
(c) 0.005 min -1 (d) 200 min -1
96. An example of a pseudo-unimoleuclar
reaction is
(a)
(b)
Dissociation of hydrogen iodide
Hydrolysis of methyl acetate in dilute
solution
(c) Dissociation of phosphorus
pentuachloride
(d)
Decomposition of hydrogen peroxide
97. About half life period of a first order
reaction, which one of the following
statements is generally false
(a)
(b)
(c)
It is independent of initial concentration
It is independent of temperature
It decreases with the introduction of a
catalyst
(d) It increases with increase of
temperature
98. The activation energy of a reaction is
zero. The rate constant of this reaction
(a)
Increases with increase of temperature
(b) Decreases with an increase of
temperature
(c) Decreases with decrease of
temperature
(d)
Is independent of temperature
99. Decomposition of nitrogen pentoxide is
known to be a first order reaction 75
percent of the oxide had decomposed
in the first 24 minutes. At the end of an
hour, after the start of the reaction, the
amount of oxide left will be
(a) Nil (b) About 1%
(c) About 2% (d) About 3%
100. If the concentration of the reactants is
increased, the rate of reaction
(a)
(b)
(c)
(d)
Remains unaffected
Increases
Decreases
May increase or decrease
[MP PMT]
101. Which of the following oxides of
nitrogen will be the most stable one
→
←
(a) 2NO 2
(g) N 2
(g) + 2O 2
(g);
K=6.7 x 10 16 mol litre -1
→
←
(b) 2NO(g) N 2
(g) + O 2
(g);
K = 2.2 x 10 30 mol litre -1
→
←
(c) 2N 2
O 5
(g) 2N 2
(g) + 5O 2
(g);
K = 1.2 x 10 34 mol litre -1
(d) 2N 2
O(g) →2N 2
(g) + O 2
(g);
←
[NCERT]
(36)

AISECT TUTORIALS : CHEMISTRY : SET-6
K = 3.5 x 10 33 mol litre -1
102. The one which is unimolecular reaction
is
[MP PAT; MP PMT]
(a)
1
2HI →H 2
+ I 2
(b) N 2
O 5
→N 2
O 4
+ O
2 2
(c) H 2
+ Cl 2
→2HCl(d) PCl 3
+ Cl 2
→PCl 5
103. The rate law for reaction A+2B = C +
2D will be
(a)
(c)
Rate = K [A][B] (b) Rate= K [A][2B]
Rate = K[A][B] 2 (d) Rate =K [C][D]2
[A][B] 2
104. A reaction rate constant is given by
(a)
(b)
(c)
(d)
k=1.2 x 10 14 e -(25000/RT) sec -1
I means
[MP PET]
log k versus log T will give a straight
line with slope as - 25000
log k versus T will give a straight line
with slope as - 25000
log k versus log 1/T will give a straight
line with slope as -25000
log k versus 1/T will give a straight line
106. The reactive ability of a compound
depends on its
(a) Atomic mass (b) Molecular mass
(c) Equivalent mass (d) Active mass
107. If the surface area of the reactants
increases, then order of the reaction
(a)
(b)
(c)
(d)
Increases
Decreases
Remain constant
Sometimes increases and sometimes
decreases
108. The law of photochemical equivalence
was given by
(a) Drapper (b) Grauths
(c) Einstein (d) Labbert
109. Relation between rate constant and
temperature by Arrhenius equation is
(a)
(b)
(c)
(d)
log e
A= log e
K + Ea
RT
log K = A Ea
RT
log e
K = log e
A- Ea
RT 2
log A = RT In E a
- In K
110. If we plot a graph between log K and 1 T
by Arrhenius equation, the slope is
E a
(a) (b) +
R
E a
(c) - (d) +
2.303R
2.303R
111. Molecularity of reaction of inversion of
sugar is
(a) 3 (b) 2
(c) 1 (d) 0
112. Time required for completion of ionic
reactions in comparison to molecular
reactions is
E a
R
E a
(a) Maximum (b) Minimum
(c) Equal (d) None
113. The unit of the velocity constant in case
of zero order reaction is
(a)
Conc. x time -1 (b) Conc. -1 x time
[CPMT]
(c) Conc. -1 x time -1 (d) Conc. x (time) 2
114. For the reaction H 2
(g) + Br 2
(g) →2HBr
(g), the experimental data suggest, rate
= K [H 2
] [Br] 1/2 The molecularity and
order of the reaction are respectively
[CPMT 1994]
(37)

AISECT TUTORIALS : CHEMISTRY : SET-6
3
3
(a) 2, (b) , 3 2
2 2
(c) 1 , 1 (d) 1, 1 2
115. The rate of the reaction
CCl 3
CHO + NO →CHCl 3
+ NO + CO is given
by Rate = K [CCl 3
CHO] [NO]. If
concentration is expressed in
moles/litre, the units of K are
[MP PET]
(a) litre 2 mole -2 sec -1 (b) mole litre -1 sec -1
(c) litre mole -1 sec -1 (d) sec -1
116. For reaction 2A + B →products, the
active mass of B is kept constant and
that of A is doubled. The rate of
reaction will then
(a)
[MP PET]
Increase 2 times (b) Increase 4 times
(c) Decrease 2 times (d) Decrease 4 times
117. In a reaction 2A + B →A 2
B, the
reactant A will disappear at
(a)
(b)
(c)
(d)
Half the rate that B will decrease
The same rate that B will decrease
Twice the rate that B will decrease
The same rate that A 2
B will form
[MP PET]
118. The minimum energy required for
molecules to enter into the reaction is
called
(a)
(c)
[KCET; EAMCET; MP PMT; MP PET]
Potential energy (b) Kinetic energy
Nuclear energy (d) Activation energy
119. The half life of a first order reaction is
10 minutes. If initial amount is 0.08
mol/litre and concentration at some
instant is 0.01 mol/litre, then t=
(a) 10 minutes (b) 30 minutes
[Roorkee]
(c) 20 minutes (d) 40 minutes
120. For an endothermic reaction, where ∆H
represents the enthalpy of the reaction
in kJ/mol,the minimum value for the
energy of activation will be
(a) Less than ∆H (b) Zero
(c) More than ∆H (d) Equal to ∆H
[IIT]
121. The rate constant (K') of one reaction is
double of the rate constant (K " ) of
anoter reaction. Then the relationship
between the corresponding activation
energies of the two reactions (E a
' and
E a
'') will be
(a) E a
'> E a
'' (b) E a
' = E a
''
(c) E a
'< E a
'' (d) E a
' = 4E a
''
[MP PET]
122. Half life period of second order reaction
is
(a)
(b)
[MP PMT]
Proportional to the initial concentration
of reactants
Independent of the initial concentration
of reactants
(c) Inversely proportional to initial
concentration of reactants
(d)
Inversely proportional to square of initial
concentration of reactants
123. In a reaction involving hydrolysis of an
organic chloride in presence of large
excess of water
RCl + H 2
O
→
ROH + HCl
[MP PET]
(38)

AISECT TUTORIALS : CHEMISTRY : SET-6
(a)
Molecularity is 2, order of reaction is
also 2
(b) Molecularity is 2, order of reaction is 1
(c) Moleuclarity is 1, order of reaction is 2
(d)
Molecularity is 1, order of reaction is
also 1
124. The thermal decomposition of a
compound is of first order. If a sample
of the compound decomposes 50% in
120 minutes, in what time will it undergo
90% decomposition
[MP PET]
(a) Nearly 240 minutes(b) Nearly 480
minutes
(c) Nearly 450 minutes (d) Nearly 400
minutes
125. The order of a reaction with rate equals
3/2 1/2
kC A CB
is
(a) 2 (b) 1
1
(c) - (d)
2
⎛
3
2
[MP PET]
126. The term
⎝ −dc in a rate equation
dt ⎠
refers to the
(a)
Concentration of the reactant
⎞
[MP PMT]
(b) Decrease in concentration of the
reactant with time
(c) Increase in concentration of the
reactant with time
(d)
Velocity constant of the reaction
127. If the rate expression for a chemical
reaction is given by Rate = k [A] m [B] n
(a)
The order of the reaction is m
[MP PMT]
(b)
(c)
(d)
The order of the reaction is n
The order of the reaction is m + n
The order of the reaction is m - n
128. Point out the wrong statement :
For a first order reaction
(a) Time for half-change (t 1/2
) is
independent of initial concentration
(b)
Change in the concentration unit does
not change the rate constant (K)
(c) Time for half-change x rate constant =
0.693
(d) The unit of K is mole -1 min -1
129. Which of the following plots is in
accordance with the Arrhenius equation
Fig. 4
130. The Arrhenius equation expressing the
effect of temperature on the rate
constant of a reaction is
(a) k = e -E a
/RT (b) k =E a /RT
[PM PET]
-E /RT
(c) k = log (d) k =Ae
RT
a
131. The half-life period of a first order
reaction is 100 sec. The rate constant
of the reaction is
e Ea
[MP PMT]
(a) 6.93 x 10 -3 sec -1 (b) 6.93 x 10 -4 sec -1
(39)

AISECT TUTORIALS : CHEMISTRY : SET-6
(c) 0.693 sec -1 (d) 69.3 sec -1
132. For the first order reaction with rate
constant k, which expression gives the
half-life period ? (Initial concentration =
a)
[MP PET / PMT]
In
(a)
2
1
(b)
k
ka
0.693/
3
(c)
(d)
ka
2ka 2
133. The rate constant is given by the
equation k = pZe -E/RT . Which factor
should register a decrease for the
reaction to proceed more rapidly ?
(a) T (b) Z
(c) E (d) P
[MP PET/ PMT]
134. The rate constant of a first order
reaction whose half-life is 480 seconds,
is
(a) 2.88 x 10 -3 sec -1 (b) 1.44 x 10 -3 sec -1
[MP PET]
(c) 1.44 sec -1 (d) 0.72 x 10 -3 sec -1
135. The reaction 2FeCl 3
+ SnCl 2
→2FeCl 2
+ SnCl 4
is an example of
(a)
(b)
(c)
(d)
First order reaction
Second order reaction
Third order reaction
None of these
[CBSE 1996; MP PET 1999]
136. If reaction between A and B to give C
shows first order kinetics in A and
second order in B, the rate equation can
be written as
[MP PET]
(a) Rate = k [A] [B] 1/2 (b) Rate = k [A] 1/2 [B]
(c) Rate= k [A] [B] 2 (d) Rate = k [A] 2 [B]
137. The following statements(s) is (are)
correct
(a)
(b)
(c)
A plot of log K p
versus 1/T is linear
[IIT]
a plot of log [X] versus time is linear for
a first order reaction X P
→
a plot of log P versus 1/T is linear at
constant volume
(d) A plot of P versus 1/V is linear at
constant temperature
138. For a first order reaction
(a)
(b)
(c)
[IIT]
The degree of dissociation is equal to
(1-e -kt )
A plot of reciprocal concentration of the
reactant vs time gives a straight line
The time taken for the completion of
75% reaction is thrice the t 1/2
of the
reaction
(d) The pre-exponential factor in the
Arrhenius equation has the dimension
of time T -1
139. For the reaction A →B, the rate law
expression is : Rate = k [A]
(a)
(b)
(c)
Which of the following statements is
incorrect
[Punjab PMT]
The reaction is said to follow first order
kinetics
The half life of the reaction will depend
on the initial concentration of the
reactant
k is constant for the reaction at a
constant temperature
(40)

AISECT TUTORIALS : CHEMISTRY : SET-6
(d)
The rate law provides a simple way of
predicting the concentration of
reactants and products at any time after
the start of the reaction
141. If initial concentration is reduced to its
1/4th in a zero order reaction,m the time
taken for half of the reaction to
complete
140. Activation energy of a chemical reaction
can be determined by
(a)
(b)
Changing concentration of reactants
[CBSE]
Evaluating rate constant at two different
temperatures
(c) Evaluating rate constants at two
different temperature
(a)
Remains same (b) Becomes 4 times
[BHU]
(c) Becomes one-fourth (d) Doubles
142. For the reaction A →B, the rate
increases by a factor of 2.25 when the
concentration of A is increased by
1.5.What is the order of the reaction
[Karnataka CET]
(d)
Evaluating velocities of reaction at two
different temperatures
(a) 3 (b) 0
(c) 2 (d) 1
(41)

AISECT TUTORIALS : CHEMISTRY : SET-6
ANSWER SHEET
Q.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans
1 B 27 B 53 B 79 D 105 C 131 A
2 D 28 A 54 C 80 D 106 D 132 A
3 B 29 B 55 C 81 B 107 D 133 C
4 C 30 A 56 C 82 B 108 C 134 B
5 D 31 C 57 A 83 B 109 A 135 C
6 B 32 B 58 B 84 A 110 C 136 C
7 C 33 A 59 A 85 B 111 B 137 A,B,C
8 C 34 C 60 C 86 A 112 B 138 A,D
9 C 35 D 61 C 87 B 113 A 139 B
10 C 36 C 62 B 88 C 114 A 140 C
11 B 37 C 63 B 89 B 115 C 141 C
12 D 38 C 64 C 90 D 116 B 142 C
13 A 39 A 65 D 91 C 117 C 143
14 C 40 C 66 B 92 C 118 D 144
15 A 41 B 67 B 93 C 119 B 145
16 A 42 C 68 A 94 D 120 C 146
17 B 43 B 69 C 95 C 121 C 147
18 B 44 D 70 B 96 B 122 C 148
19 B 45 B 71 B 97 D 123 B 149
20 A 46 D 72 B 98 D 124 D 150
21 D 47 C 73 A 99 D 125 B 151
22 A 48 B 74 A 100 B 126 B 152
23 B 49 B 75 D 101 A 127 C 153
24 D 50 A 76 B 102 B 128 D 154
25 B 51 B 77 A 103 C 129 A 155
26 B 52 C 78 D 104 D 130 D 156
(42)

AISECT TUTORIALS : CHEMISTRY : SET-6
CHAPTER-2
NUCLEAR CHEMISTRY (RADIOACTIVITY)
Nuclear Chemistry (Radioactivity)
1. Nucleus
Nucleus is found to be a source of
tremendous amount of energy which has
been utilised for the destructive as well as
constructive proposes. Hence the study of
nucleus of an atom has become so important
that it is given a separate branch of chemistry
under the heading of nuclear chemistry.
The nucleus occupies a central place in
the atom. Nearly all the mass of an
atom is concentrated in the nucleus.
The nucleus is nearly 10 4 times smaller
in size and 10 12 times smaller in
volumes than the atom. The empirical
relationship between the size of the
nucleus and its mass number is
R = R 0
A 1/3
where R = Radius of the nucleus with
mass number A
Ro = A constant = 1.4 x 10 -13 cm.
Since nucleus has a radius of only
about 10 -15 m, the density of nucleus is
enormous, more than 10 12 times that of
lead. It is of the order of 10 14 g/c.c.
Although about 20 different particles
have been detected as the products of
different nuclear reactions, it does not
mean that all these particles are
present in the nucleus. Actually nucleus
in believed to consist of two building
blocks, protons and neutrons, which are
collectively called nucleons. Other
prticles* are considered as created by
stresses in which energy is converted
into mass or vice versa, e.g. an electron
( β-particle) from a radioactive nucleus
may be regarded as derived from a
neutron in the following way.
Neutron →Proton + Electron
Similarly, photons are produced from
internal stresses within the nucleus.
2. Nuclear Forces.
Since the radius of nucleus is very small
~ 10 -15 m, two protons lying in the
nucleus are found to repel each other
with an electrostatic force of about 6
tonnes. Now since the radius of the
nucleus is of the order of 10 -15 m, the
question arises how such a large
number of protons are present in such a
tiny space without repulsion.
It is postulated that stronger
proton-neutron, neutron-neutron and
even proton-proton attractive forces
exist in the nucleus. These attracitive
forces are called nuclear forces.
Unlike electrostatic forces which
operate over long ranges, the nuclear
forces operate only within small
distance of about 1 x 10 -15 m ( 1 fermi**)
and drop rapidly to zero at a distance of
1.4 x 10 -13 m. Hence these are referred
to as short range forces. Nuclear forces
are nearly 10 21 times stronger than
electrostatic forces
(43)

AISECT TUTORIALS : CHEMISTRY : SET-6
3. Nuclear Stability.
The stability of nucleus may be
discussed in terms of any one of the
following :
3.1 Mass defect and nuclear binding
energy. The most acceptable theory
about the atomic nuclear stability is
based upon the fact that the observed
atomic mass of all known isotopes
(except hydrogen) is always less from
the sum of the weights of protons and
neutrons (nucleons) and electrons
present in it. Consider the case of
helium having 2 protons, 2 neutrons
and 2 electrons.
Mass of 2 electrons = 2 x 0.000548 =
0.001086 amu
Mass of 2 protons = 2 x 1.00758 =
2.01516 amu
Mass of 2 neutrons = 2 x 1.00893 =
2.01786 amu
∴Expected mass of helium = 4.03410
amu
However, the actual mass of helium
= 4.00390 amu
∴Difference between the expected
∆mass and actual mass of helium, i.e.
m = 4.03410 - 4.00390 amu
= 0.03020 amu
The difference between the expected
mass (calculated by adding the masses
of protons, neutrons and electrons
present )and the actual mass of an
isotope is called mass defect; it is
denoted by m. ∆
The natural question is that where this
mass has gone ? It has been suggested
that this mass is converted into energy
which is released in the formation of the
given nucleus from individual protons
and neurtrons. The release of energy
results in the stability of the nucleus, i.e.
it helps in binding the nucleons together
and hence is called binding energy.
Thus binding energy of a nucleus may
be defined as the amount of energy
released during the formation of
hypothetical nucleus from its protons
and neutrons. Obviously, the same
amount of energy will be required to
separate the nucleons apart. Hence
binding energy of a nucleus may also
be defined as the energy required to
disrupt it into its constituent protons and
neutrons.
Mathematically, binding enrgy can be
calculated from Einstein equation,
E = ∆mc 2
where ∆m is the mass lost, i.e. mass
defect c is the velocity of light (3 x 10 10
cm/sec)
Alternatively, it may simply be obtained by
multiplying the mass defect with 931.5 MeV
i.e.
Binding energy = ∆M x 931.5 Mev
This is because 1 amu* = 931.5 meV
Thus the binding energy of helium nucleus
= ∆ m x 931.5 MeV
= 0.0302 x 931.5 MeV
= 28.14 MeV
If ∆ m is in gram E will be in MeV per gm
atom
Evidently greater the mass defect, greater
is the binding energy of the nucleus.
(44)

The binding energy of a nucleus when divided
by the number of nucleons gives the mean
binding energy per nucleon. Thus the mean
binding energy per nucleon of helium
(having 4 nucleons ).
= 28.14
4
= 7.03 MeV
The binding energy per nucleon is a measure
of the stability at the nucleus. The greater the
binding energy per nucleon, the more stable
is the nucleus.
(i)
(ii)
(iii)
The averages binding energy for most
of the nuclei is in the vicinity of 8 MeV.
Nuclei having binding energy per
nucleon very near to 8 MeV are more or
less stable.
Iron has the maximum average binding
energy (8.79 MeV) and thus its nucleus
is thermodynamically most stable.
The isotopes with intermediate mass
number 40 to 100 are most stable. The
elements with low mass numbers or
high mass numbers tend to become
stable by acquiring intermediate mass
number. Evidently, nuclei of lighter
elements combine together to form a
heavier nucleus of intermediate mass
number (nuclear fusion); while the
nuclei of heavy elements split into two
lighter nuclei of intermediate mass
number (nuclear fission). In either case
energy is relased and hence the stability
is enhanced.
Relative stability of isotopes and binding
energy. Value of binding energy predicts the
relative stability of the different isotopes of an
element. If the value of binding energy is
negative, the product nucleus or nuclei will be
less stable than the reactant nucleus or nuclei
AISECT TUTORIALS : CHEMISTRY : SET-6
(45)
but if the binding energy is positive, the
product nucleus is more stable than the
reactant nucleus. Thus the relative stability of
the different isotopes of an element can be
predicted by the values of binding energy for
each successive additionof one neutron to
the nucleus.
2He 3 + o n1 → 2 He 4 + 20.5 MeV
2He 4 + o n1 → 2 He 5 − 0.8 MeV
Therefore 2He 4 is more stable than 2He 3
and 2 He 5 .
3.2. Packing fraction. The difference of
actual isotopic mass. and the mass number
in terms of packing fraction is defined as
Packing fraction
Actutal isotopic mass − Mass number
= x10 4
Mass number
The value of packing fraction depends upon
the manner of packing of the nucleons within
the nucleus. Its value can be negative,
positive or even zero.
Carbon 12 has zero packing fraction because it
is taken as a reference on the atomic scale
and its actual isotopic mass (12) is equal to
its mass number (12).
The negative or zero value of the packing
fraction means that the actual isotopic mass
is less or equal to the mass number. This in
turn indicates that some mass has been
transformed into energy (binding energy)
during formation of nucleus. Such nuclei are
therefore more stable.
The positive packing fraction should imply the
opposite i.e. the nuclei of such isotopes
should be unstable. However this
generalisation is not strictly correct especially
for elements of low mass number. For these

AISECT TUTORIALS : CHEMISTRY : SET-6
elements, though packing fraction is positive,
yet they are stable. This is explained on the
basis that the actual masses of protons and
neutrons (of which the nuclei are composed)
are slightly greater than unity.
In general, lower the packing fraction, greater
is the binding energy per nucleon and hence
greater is the stability. The relatively low
packing fraction of he, C and O implies their
exceptional stability. Packing fractrion is least
for Fe( negative) and highest for H (+78).
3.3. Meson theory of nuclear forces.
Neutron is found to play a leading role in
binding the nuclear particles. It has been
established that neutron proton attractions
are stronger than the proton-proton or
neutron-neutron attractions. This is evident by
the fact that the deuteron ,H 2 having one
proton and one neutron is quite stable while
no particle having either two neutrons or two
protons is known.
Yukawa in 1935 put forward a postulate that
neutrons and protons are held together by
very rapid exchange of nuclear particles
called pi mesons. (cf the formation of a
covalent bond by sharing of electrons.). The
nuclear forces called into play by this rapid
exchange of pimesons between nucleons are
also called exchange forces.
The binding forces between unlike nucleons
(p and n) are explained by the oscillation of a
charged pi meson ( π + or π − )
(a) p1 + n2⇌n1+ π + + n 2
⇌ n 1
+ p 2
(b) p1 + n2 ⇌ p1 + λ - + p2 ⇌ n 1
+ p 2
Binding forces between like particles (p-p) or
n-n) result from the exchange of neutral
mesons (l) as represented below.
(c) p1 + p2 = λ o or p1 + π o = p2
(d) n1 = n2 +λ o or n1 + λ o ⇌ n2
3.4. Neutron-proton ratio and nuclear
stability. The nuclear stability is found to be
related to the neutron/proton (n/p) ratio. If for
different elements the number of neutrons is
plotted against the number of protons, it is
found that the elements with stable nuclei
(non radioactive elements) lie within a region
(belt) known as zone or belt of stability
Fig. 5
In Figure following points are worth noting
(i)
For elements with low atomic number
(less than 20), n/p ratio is 1, i.e. the
number of protons is equal to the
number of neutrons. Remember that
n/p ratio of 1
H 1 is zero as it has no
(46)

(ii)
(iii)
neutron. Nuclide with highest n.p ratio is
H 3 (n/p =2.0).
With the increase in atomic number
although the number of protons
increased but the number of neutrons
increased much more than the number
of protons with the result the n/p ratrio
goes on increasing from 1 till it
becomes nearly equal to 1.5 at the
upper end of the belt.
When the n/p ratio exceeds 1.52 as in
elements with atomic number 84 or
higher, the elements becomes
radioactive and undergoes
disintegration spontaneously. Note that
these elements lie outside the zone of
stability.
The way an unstable nucleus disintegrates is
decided by its position with respect to the
actual n/p plot of stable nuclei ( the zone of
stability).
(a) Neutrons to proton (n/p) ratio too high.
If the n/p ratio is too high i.e. when the
nucleus contains too many neutrons it falls
above the zone of stability. The isotope would
be unstable and would tend to come within
the stability zone by the emission of a
β-ray(electron) Electron, in turn is produced in
the nucleus probably by the following type of
decay of a neutron.
on 1 → 1 H 1 + −1e 0 (beta particle)
The electron thus produced is emitted as a β
particle and thus the neutron decay ultimately
increased the number of protons, with the
result the n/p ratio decreases and comes to
the stable belt.
Consider the example of C 12 and C 14 . In C 12
the n/p ratio (6/6 ) is 1; hence its nucleus is
AISECT TUTORIALS : CHEMISTRY : SET-6
(47)
quite stable. On the other hand in C 14 the n/p
ratio(8/6) is 1.3; hence it should be unstable.
In practice also it is found to be so and C 14
decays in the following way to give N 14 (n/p
ratio =1).
6C 14 → 7 N 14 + (−1e 0 )
( n p = 8 6 = 1.33) ( n p = 7 7 = 1.0)
Similarly,
11Na 24 → 12Mg 24 + − 1 e0
( n p = 13 = 1.18) ( n 11 p = 12 = 1.0)
12
92U 238 → 90 Th 234 + 2 H e4
( n p = 146 = 1.587) ( n 92 p = 144 = 1.6)
90
(b)
Neutron to proton ratio (n/p) too low,
i.e. when the nucleus contains excess
protons. There are no naturally
occurring nuclides with n/p ratio less
than 1, however there are many
artifically nuclides. In such cases the
nucleus lies below the zone of stability it
would again be unstable and would tend
to come within the zone of stability by
losing a positron.
6C 11 → 5B 11 + 1 e0
( n p = 5 6 = 0.83) ( n p = 6 5 = 1.2)
7C 13 → 6B 13 + 1 e0
( n p = 6 7 = 0.86) ( n p = 7 6 = 1.16)
3.5. Nuclear shell model. According to this
theory, nucleus of atom, like extra nuclear
electrons, also has definite energy levels
(shells). The shell structure is supported by
the existence of periodicity in the nuclear
properties. For example elements with even
number of protons and neutrons are more
abundant more stable and richer in isotopes.
Nuclides with odd number of protons and
netrons are least abundant is nature (only 5
are known 1
H 2 , 3Li 6 , 5
B 10 , 7
N 14 and 73Ta 180 ).

AISECT TUTORIALS : CHEMISTRY : SET-6
Thus elements have a tendency to have even
number of both protons and neutrons. This
suggests that like electrons, nucleon particles
in the nucleus are paired. Magnetic fields
ofthe two paired protons spinning in opposite
direction cancel each other and develop
attractive forces which are sufficient to
stablize the nucleus.
Further nuclei with 2, 8, 20, 28, 50, 82 or 126
protons or neutrons have been found to be
particularly stable with a large number of
isotopes. These numbers, commonly known
as magic numbers are defined as the
number of nucleons required for completion
of the energy levels of the nucleus. Nucleons
are arranged in shells as two protons or two
neutrons (with paired spins) just like electrons
arranged in the extra-nuclear part. Thus the
following nuclei
2 He4, 8O 16 , 20
Ca 40 and 82
Pb 208
containing protons 2,8,20 and 82 respectively
(all magic numbers) and neutrons 2,8,20 and
126 respectively (all magic numbers) are the
most stable.
Magic numbers for protons :
2, 8, 20, 28, 50, 82 , 114
Magic numbers for neutrons :
2, 8, 20, 28, 50, 126, 184, 196
The relatively higher stability of 82
Pb 208 can be
explained by magic numbers (magic numbers
for protons = 82, magic numberfor neutron=
126).
Nuclei with nucleons just above the magic
numbers are less stable and hence these
may emit some particles to attain magic
numbers.
Illustration 1. If the atomic masses of
lithium, helium and proton are 7.01823 amu,
4.00387 amu and 1.00815 amu respectivey,
calculate the energy that will be evolved in the
reaction,
Li 7 + H 1 → 2 He 4 + energy
Given that 1 amu = 931 MeV
Solution Total mass of the reacting species
(Li7 and H1)
= 7.01823 + 1.00815 = 8.02638 amu
The mass of the resulting species (2 He4)
= 2 x 4.00387 = 8.00774 amu
Mass of reacting species converted into
energy e.e.
∆M = 8.02638- 8.00774 = 0.01864 amu
... Energy evolved in the reaction
= 0.01864 x 931 = 17.354 MeV
Illustratiion 2. Calculate the mass defect and
binding energy per nucleon for 27CO59. The
mass of CO59 = 58.95 amu. mass of
hydrogen atom = 1.008142 amu and mass of
neutron = 1.008982 amu].
Solution Number of protons in 27
CO 59 = 27
. . . Number of neutrons = 59-27=32
Dm = (1.008142 x 27 + 1.008982 x 32) -
58.95 = 0.556438 amu
The binding energy (E. B
. per nucleon)
=
∆M x931
Mass of cobalt
=
= 8.77 MeV
0.556438 x 931
59
MeV
Illustration 3. Calculate the energy evolved
(in joules ) per mole of hellium formed by he
fusion of two deuterum nuclei. The masses of
deuterium and hellium nuclei are 2.014 amu
and 4.00 amu respectively.
(48)

AISECT TUTORIALS : CHEMISTRY : SET-6
Solution. The mass of two deuterium nuclei
= 2 x 2.014 amu = 4.028 amu
Total mass of one helium nuclei formed
= 4.00 amu
. . . DM = 4.028 - 4.00 = 0.028 amu
Hence the energy (per atom) evolved
= 0.028 x 1.66 x 10 -27 x (3 x10 8 )2 J
= 4.183 x 10 -12 J
Energy evolved per mole of He formed
= 4.183 x 10 -12 x 6.023 x 10 23 J
= 2.521 x 10 12 J
Illustration 4. Calculate the loss in mas
accompanying combusion of one mole of the
given fuel which yeilds 900 kJ of heat energy.
Solution . Here DE = 900 kJ = 900 x 103 J,
c= 3 x 10 8 m sec-1
Now we know that DE = Dmc 2
. . . ∆m = ∆E
C 2
=
900 x103
(3 x 10 8 ) = 1 x 10 −12 kg
2
Illustration 5. Binding energy per nucleon of
a nuclide with mass number 102 is 8.25 MeV.
Calculate the mass defect.
Solution Total binding energy
= Binding energy per nucleon x No. of
nucleons
= 8.25 x 102
= 841.5 MeV
Now we know that
Binding energy = Mass defect x 931.5
Binding energy
Mass defect =
931.5
= 841.5 = 0.9034 amu
931.5
Illustration 6 Calculate the packing fraction
of Ar 40 (isotopic weight of Ar = 39.96238)
Solution Isotopic atomic weight of Ar 40 =
39.96238
Mass number of argon 40 = 40.00
. . 39.96238 − 40
. Packing fraction = x 10 4
40
0.03762 x 104
= = − 9.405
40
Illustration 7 Of the following four nuclei.
2 He4 , 6
C 12 , 7
N 15 and 4
Be 7
which is expected to acquire radioactive
properties ?
Solution Since n/p ratio in He, C, N and Be is
1,1,1.14 and 0.75 the last one i.e. 4 Be7
differs to a great extent from 1 therefore it
may acquire radioactive property.
4. Nuclear Reactions.
We know that in a chemical reaction, only
electrons (Extra-nuclear part) of the atom
take part while the nucleus of the atom
remains unaffected. However, the reverse
reactions (i.e., where only nuclei of atoms
take part in reactions) are also possible. Such
reactions in which nucleus of an atom itself
undergoes spontaneous change or interact
with other nuclei of lighter particles resulting
new nuclei and one or more lighter particles
are called nuclear reactions.
4.1 Important features of nuclear
reactions.
(i)
(ii)
Nuclear reactions are written like a
chemcial reaction. As in a chemical
reaction, reactants in a nuclear reaction
are written on the left hand side and
products on the right hand side with an
arrow in between them.
Mass number and atomic number of
the elements are written in a nuclear
reactins. Mass number and atomic
(49)

AISECT TUTORIALS : CHEMISTRY : SET-6
(iii)
(iv)
(v)
(vi)
number of the element involved in a
nuclear reaction are inserted as supers
cripts and subscripts respectively on the
symbol of the element for example
27
AI, 27
Al or 13
A 27 stands for an atom
13 13
of aluminium with mass number 27 and
atomic number 13.
Mass number and atomic number are
conserved. In a nuclear reaction the
total mass numbers and total atomic
numbers are balanced on the two sides
of the reaction (recall that in an ordinary
reactions the total number of atoms of
the various elements are balanced on
the two sides).
Energy involved in the nuclear
reactions is indicated in the product
as + Q or -Q for reactions accompained
by release or absorption of energy
respevtively.
Important projectiles are α particles
( 2
He 4 ) proton ( 1
H 1 or p), deuteron ( 1
H 2 or
1 D2 ), neutron ( o
n 1 ), electron (β-particle,
-1e 0 or e) and positron (+1e o ).
Representation of nuclear reactions.
For example
7 N14 + 2
He 4 → 8
O 17 + 1
H 1 + Q
Sometimes a short hand notation is used,
e.g the above reaction can be represented as
below.
7 N14 (α, p) 8
O 17
4.2 Nuclear reactions vs Chemical
Reactions. Nuclear reactions differ from
chemical reactions in the following respects.
(i)
As per definition chemical reactions
depends upon the number of
extranuclear electrons while nuclear
(50)
(ii)
(iii)
(iv)
(v)
reactions are independent upon the
electrons but depend upon the nature of
the nucleus.
Chemical reactions involves some loss
gain or overlap of outer orbital electrons
of the two rectant atoms. On the other
hand, nuclear reactions involve
emission of some light particles (α-,β-,
positron etc.) from the nucleus of the
atom to form another element.
The chemical reactivity of the element
is dependent on the nature of the bond
present in the concerned compound.
On the other hand, the nuclear reactivity
of the element is independent of its
state of chemical combination e.g.
radium whether present as such or in
the form of its compound shows similar
radioactivity.
The energy change occurring in nuclear
reactions is very high as compared to
that in chemical reaction. Again in
chemical reactions the energy is
expressed in kcal per mole while in
nuclear reactions the energy is
expressed in MeV per nucleus. Nuclear
reactions which liberate energy are
called exoergic reactions and which
absorb energy are called endoergic
A chemical reaction is balanced in
terms of mass only while a nuclear
reaction must be balanced in terms of
both mass and energy. In endoergic
reactions, the mass of products is more
than the mass of reactants while in
exoergic reactions the mass of products
is less than the mass of reactants.

AISECT TUTORIALS : CHEMISTRY : SET-6
(vi) The chemical reactions are dependent
on temperature and pressure while the
nuclear reactions are independent of
external conditions.
5. Types of Nuclear Reactions
There are five importants types of nuclar
reactions which are expected to take place
when a projectile strikes the nucleus of an
atom.
(a) Projectile Capture Reactions -When
projectile is captured without emission of any
particle, For example,
12
6 C+ 1 1 H → 13 7 N + γ − rays
(b) Projectile capture particle Emission
Reactions: When projectile is captured with
the emission of one or more particles with the
formation of a stable nucleus. The nature of
emitted particle depends upon the energy of
the projectile for example
14
7 N + 1 0 n → 14 6 C + 1 1 H
Neutron
11
5 B + 1 1 H → 11 6 C + 1 0 n
Proton
Proton
Neutron
27
13AI + 1 1 H → 24 12 Mg + 4 2 He
Proton
10
5 B + 4 2 He → 13 7 N + 1 0 n
α-particle
α-particle
Neutron
(c) Fission Reactions - When nucleus
breaks up into two or more fragments. For
example,
235
92 U + 1 0 n → 141 56 Ba + 92 36 Kr + 2 − 3 1 0 n + energy
(d) Fusion Reactions - Where nuclei fuse
into one another to form a bigger nucleus.
For example
2
1H+ 3 1 H → 4 2 He + 1 0 n + 17.6 MeV Energy
(e) Spallation Reactions - Where high
speed projectile chops off a fragment of the
nucleus leaving behind a smaller nucleus. For
Example
63
29Cu + 4 2 He(400MeV) → 37 17 CI + 14 1 1 H + 16 1 0 n
High energy Protons
Neutrons
However it is more useful to classify
nuclear reactions according to the particle
used as projectile. The common energetic
particles used as projectile are α Particle
( 4 He), proton 2 (1 H), Neutron 1 (1 0n) and
deuteron ( 2 1H). When these particles are
used as projectile, the absorption of the
projectile by the nucleus caused the neutron
to proton ratio of the nucleus to disturb. The
resulting compound nucleus, therefore, ejects
some particle in order to bring the ratio of
neutron to proton to a stable one. The ratio of
neutron to proton lowers down, if a neutron of
β particle is emitted. The ejection of a proton
or position results in the increase of neutron
to proton ratio. Some examples of nuclear
reactions induced by these particles are given
below :
23
11Na + 1 1 H → 23 12 Mg + 1 0 n [ n pdecreases]
Proton
Neutron
15
7 N + 1 1 H → 12 6 C + 4 2 He [ n p decreases]
Proton
α-particle
12
6 C + 2 1 H → 10 5 B + 4 2 He [ n p Cons tan t]
Deuteron
α-particle
19
9 F + 4 2 He → 22 10 Ne + 1 1 H [ n pIncreases]
α-particle
Proton
27
13AI + 1 0 n → 24 11 Na + 4 2 He [ n p Increases]
Neutron
α-particle
(51)

AISECT TUTORIALS : CHEMISTRY : SET-6
Nuclear reactions can also be classified into
two classes based on spontaneous
distinegration or artificial distengration (a)
Those nuclear reactions in which nucleus of
heavy atoms spontaneously disnegration with
the emission of radiations. In such reactions
ony one nucleus is involved. natural
radioactivity is an example of such reactions
(b) Those nuclear reactions in which a
relatively heavier nucleus is bombarded with
a lighter nuclei. The lighter nuclei is usually
an energetic particle, such a neutron, proton,
electron deutron or α particle etc. Such
reactions are covered under the general title
aftificial radioactivity and artifical
transmutation.
1
10
The products in a nuclear reaction are a
heavy nucleus and one of the following :
(1) Proton (2) Neutron (3) Alpha particle (4)
Beta particle or Electron (5) Deutron (6)
Position or (7) γ − rays.
The heavy nucleus may or may not be stable.
If it is stable, no more radioactivity is
expected to occur, but if it is unstable, it may
undergo distintegration spontaneosuly itself
like natural radioactive elements. This is
called induced radioactivity.
There are certain reactions in which a heavy
nucleus is rendered unstable by bombarding
it with neutrons. As a result, the nucleus
breaks up into two parts of comparable
masses emitting several neutrons and a huge
amount of energy. Such nuclear reactions
are called nuclear fission.
Few nuclear reactions are also known
where isotopes of very little element, such as
deuterium etc many react with one another to
form heavier or more stable nucleus. Very
high temperature of millions of deress, is
required to cause such reaction, which are
called fusion reactions.
6. Q VALUE OF NUCLEAR
REACTIONS
The energy change Q accompanying a
nuclear reaction should also be written with
the complete nuclear reaction. For example
14
7 N + 4 2 He → 17 8 O + 1 1 H + Q
Where Q is called nuclear reaction energy.
Just like chemical reaction,m the value of
Q may be negative or positive depending
upon whether the reaction is exoergic or
endoergic(exthermic or endothermic in the
chemical reactions. ) the value of Q may be
calculated as follows :
Sum of masses = 14.0031 + 4.0026 =18.0057
or Reactants
a.m.u.
Sum of masses = 16.991 + 1.0078 =18.0069
or Products
Change in
mass
a.m.u.
= 18.0069 -18.0057 =0.0012
a.m.u.
Since the above reaction is accompanied by
an increase in mass of 0.0012 a.mu.an
equivalent amount of energy is absorbed in
this case and hence the value of Q is Positive
and the reaction is therefore, endoergic.
Since 1 a.m.u. = 931.5 MeV, hence
Q=0.0012 x 931.5 = 1.118 MeV. In the
endoergic reaction, the total mass of the
products is more than that of reactants.
whereas in exoergic reaction, the total
mass of products will be less than that of
the reactants.
Examples (a) Endoergic Reactions :
12
6 C + 2 1 H → 13 7 N + 1 0 n − 0.2844MeV
(52)

AISECT TUTORIALS : CHEMISTRY : SET-6
14
7 N + 4 2 He → 17 8 O + 1 1 H − 1.118 MeV
(b) Exoergic Reactions :
10
5 B + 1 0 n → 7 3 Li + 4 2 He + 2.784 MeV
n 235 92 U + 1 0 n → 143 56 Ba + 90 36 Kr + 3 1 0 n + 200Mev
7. Fundamental Particles
After the discovery of nucleus by
Rutherford, it was shown that nucleus is
composed of protons and neutrons, but after
ther discovery of sub-atomic or
fundamental particles. the nuclear picture
has grown quite entangled and obscure
today. The existence of more than 30 types of
nuclear or subatomic particles has been
established so far. Some of these particles
are stable, while others are unstable. The
stable particles are electron, proton,
antiproton and positron (all mass particles)
and photon, neutrino etc. (energy particles).
Unstable fundamental particles include
neutron, meson. V particles etc. It should be
noted that the term 'stable' denotes an
absence of decay but does not necessarily
imply long life. For example position is a
stable particle, but usually short lived,
because it is readily destroyed by interaction
with an electron. It should also be noted that
all these fundamental particles have been
detected in the products of various nuclear
reactions but nucleus is not necessarily be
a mixture of all these fundamental
particles. For example photons can never
be present in the nucleus. The mass
particles such as electrons, positrons and
mesons are also not expected to exist as
components of nucleus, but these are
created by stresses in which energy is
converted into mass.
Electron was first discovered by J.J.
Thomson (1887) who showed that it is a
universal constituent of matter. The change
on an electron is 4.8024 x 10 -10 e.s.u. or 1.602
x 10 -19 Coulombs. Its mass is.
Ilustration 8. 4
Be 7 captures a K-electron into
its nucleus. What will be the mass number
and atomic number of the nuclide formed ?
Solution When a nucleus captures a
K-electron, a proton is converted to neutron.
So the mass number does not change but the
atomic number reduces by 1 unit. Thus the
mass number and atomic number of the
resulting nuclide will be 7 and 3 respectively.
Illustration -Complete the following nuclear
reactions
(i) 11 Na 23 + 1 H 1 → 12 Mg 23 + ?
(ii) 93 Np 239 → 94 Pu 239 + ?
(iii) 92 U 238 → 90 Th 234 + ?
(iv) 13 Ai 27 + ? → 15 P 30 + 0 n 1
Solution.
(i) 11 Na 23 + 1 H 1 → 12 Mg 23 + ?
Sum of mass numbers of reactants = 23 +
1=24
Sum of atomic mumbers of reactants =
11+1=12
Mass number of Mg (present in product) =23
At. number of Mg (present in product) = 12
. . . Mass number required in product
for balancing the equation = 24-23=1
At. Number required in product = 12-12=0
In other words, the missing particle in the
product has mass number 1 and atomic
number 0, i.e. it is
O n1
Thus
11Na 23 + 1 H 1 → 12 Mg 23 + on 1
(ii) 93 Np 239 → 94 Pu 239 + ?
(53)

AISECT TUTORIALS : CHEMISTRY : SET-6
The reaction indicates that the reactant and
product have the same mass number (239)
but the atomic number of the product is
incresed by 1 unit. Thus the missing product
corresponds to the particle having a mass
number of zero and a charge of -1 which is a
β-particle (electron).
93
Np 239 → 94 Pu 239 + −1e 0 (β − particle)
(iii) 92 U 238 → 90 Th 234 + ?
In this change there is a decrease of atomic
number by 2 units and of atomic mass by 4
units. In other words the product must have a
particle having 2 units positive charge and 4
units atomic mass which corresponds to
2He 4 (α-particle). Thus,
92U 238 → 90 Th 234 + 2 He 4
(iv) 13 AI 27 + ? → 15 P 30 + on 1
Sum of mass numbers of the products = 30
+1=31
. . . Mass number of the missing species in the
rectant
tant = 31-27=4
Sum of atomic number (positive charges) of
the products = 15 +0 = 15
. . . Positive charge (At number) of the missing
species in the reactant = 15-13=2.
The species having atomic mass of 4 units
and atomic number of 2 units is helium
(α-particle) i.e. 2
He 4 Thus
13AI 27 + 2 He 4 → 15 P 30 + on 1
8. Radioactivity
Radioactivity may be defined as a process in
which nuclei of certain elements undergo
spontaneous distinegration (Transformation
into another element by the ejection of α- or
β-particle) at a rate characterstic for each
particular active isotope (Becequerel, 1896).
All the heavy elements from bismuth (atomic
number 83) through uranium and also a few
of the lighter element possess radioacive
properties. However, the radioactive
elements differs widely e.g. radium atoms
have about three million times the activity of
uranium atoms. Uranium in the form of
potassium uranly suplhate, KUO 2
(SO 4
) 2
was
the first compound found to be radioactive.
Radioactive changes are spontaneous.
These are not controlled by temperature
pressure or nature of chemical combination.
9. Radioactive Rays (Radioactive
emanations).
Radioactive rays are characterised by the
following properties.
(i)
(ii)
(iii)
(iv)
They blacken photographic plates.
They pass through thin metal foils.
They produce ionization in gases
through which they pass.
They produce luminescence in zinc
sulphide, barium platinocyanide,
calcium tungstate, etc.
Radioactive radiations are composed of three
importants rays, namely α-,β- and γ- rays
which differ very much in their nature and
properties, eg. pentrating power, ionising
power and effect on photographic plates.
Changes in Atomic Nuclei by
Emission of α, β Particles and γ Rays
α-ray change (or α decay) The change
occurs due to the emission of an α-prticle by
an unstable nucleus, since an alpha particle
is made up of 2 protons and 2 neutrons,
emission of an alpha particle from an
unstable nucleus during radioactivity forms a
(54)

AISECT TUTORIALS : CHEMISTRY : SET-6
new element which has atomic number 2 less
and mass number 4 less than the unstable
atom. The unstable atom is called parent
nuclide and new element formed is called the
daughter element. For example a radium
atom (atomic mass= 226 and atomic number
=88 loses an α-particle to form radon (atomic
10. Comparison of the Properties of α, β and γ rays
Property α-particles β-particles γ-rays
1. Nature Helium nucleus Electron Electromagnetic
radiations of short
wavelength
2. Electrical charge +2 e units -e unit No charge (i.e. zero) or
3. Mass 4 times of H atom
electrically neutral
1
th of H atom
1837
Nil (Non material)
4. Velocity About 1/10 th of the About th to the velocity of Velocity of light 3 x 10 8
velocity of light 2 x 10 7 light 2.36 x 10 8 to 2.83 x 10 8 m/sec
m
/ m
sec
/ sec
5. Relative penelrating
1 i.e. 10 -3 cm thick Al
100 i.e. 10 -1 cm thick Al plate
10000 i.e. penetrate 10
power
plate stopped by 0.1 mm
or .1 cm thick Al plate (More
cm thick Al plate
thick Al plate (Least)
than α)
(Maximum)
6. Fluorescence on Maximum due to high
ZnS plate
kinetic energy
7. gases power for HIgh (1) due to high
speed an mass
Less than α
Low ⎛ ⎝ 1
100
⎞
⎠ due to low mass Very
Least
low ⎛ ⎝
1
10,000
⎞
⎠
8. Effect on
photographic plate
Strong effect Less than α Less than β
9. Scattering Get scattered while
passing through metal
foils
10. Kinetic energy High due to large mass
and high einetic energy
11. Toxicity They damage the
tissues only in intimate
contact (Non-toxic)
No
Low due to much lower mass
and low kinetic energy
No effect (Non-toxic)
No
Nil due to no mass
Most harmful to body or
living tissues (Toxic)
(55)

AISECT TUTORIALS : CHEMISTRY : SET-6
mass=222 and atomic number=86) as shown
below -
88Ra 226 → 86 Rn 222 + 2 He 4 ; 92 U 238 →
90Th 234 + 2 He
4
⎯⎯⎯ →
2He 4 ]
β-ray change -This change eoccurs due to
the emission of β-particle by an unstable
nucleus. Since a β−particle is just an electron
having negligble mass, but one unit of
negative charge. So the loss of a β particle
results only in the increase of unit positive
charge on the nucleus of the atom i.e. atomic
number of the element increased by one. For
example radium atomic number= 88 and
atomic weight=228) loss a particle to form
actinium (atomic weight =228 and atomic
number =89) as shown below-
88Ra 228 → 89 Ac 228 + 1 e 0 (β − particle)
γ ray change- Gamma rays are light rays
with very short wavelength. They have no
charge and no mass. Emission of gamma
rays does not change the number of protons
and neutrons. In most of the transormations
γ-rays are also emitted along with alpha and β
particles. Most of the nuclear reactions take
place in the excited state. Gamma rays are
emitted when the excited nucleus returns to
its ground state.
11. Difference between an Alpha
particle and a Helium Atom
An alpha particle in made up of 2 protons and
2 neutrons whereas a helium atom is made
up of 2 protons, 2 neutrons and 2 electrons.
Therefore an alpha particle is doubly
positive charged helium nucleus. An alpha
particle is positively charged while a helium
atom is neutral.
12. Difference Between β particles and
Electrons
β-particles
Electrons
They are emitted with They revolve round the
high speed from the nucleus in orbits and
nucleus of an atom can be removed only by
during
processes.
radioactive supplying enough
energy to the atom
When a β-particle is When an electron is
removed the atomic removed from an atom.
number of the atom Its positive charge
increases by one unit.
13. Source of β-rays
increased by one unit
and there is no effect on
the atomic number of
atom involved.
It is believed that β-particles (or electrons)
are ejected during radioactive change as a
result of neutron deacy.
Neutron → Pr oton → Electron
14. Bacquerel Rays
The emission of radiations carried out by
radioactive substances are known as
Bacquerel rays. These rays are consists of
alpha (α) beta (β) and gamma ( ) rays γ
15. Behaviour of Radioactive
Radiations When Passed Through an
Electric and a Magnetic Field
When radioactive radiations are passed
through magnetic and electric field, following
observations were made -
(i) Certain rays were readily deflected by
electric and magnetic fields in such a
direction (i.e. towards positive end) indicating
(56)

AISECT TUTORIALS : CHEMISTRY : SET-6
thereby that they carry a negative charged
and are termed as β-rays.
(ii) The deflection of α rays was far slighter
than β-rays and was in the opposite direction.
Consequently they possess a positive charge.
(iii) The γ rays showed no deflection in
electric and magnetic fields. Hence these
rays are electrically neutral electromagnetic
waves.
16. Disintegration Theory of
Radioactivity.
According to this theory put forward by
Rutherford and Soddy in 1902, radioactive
elements (which are generally heavy
elements) are unstable (the source of
unstability is nucleus) and undergo
spontaneous, break down from one chemical
atom to another. Such breakdowns are
successive and proceed with the emission of
either a helium nucleus ( 2
He 4 ) or a β-particle
(-1e 0 ). In case of natural radio-elements,
species of the atom formed in the process
are themselves generally still unstable and
therefore further undergo disintegration. This
process is continuous until finally a stable
and, therefore, non-radioactive nuclear
species results.
17. Laws of Radioacive Disintegration.
(i)
(ii)
Atoms of all radioactive elements
undergo spontaneous disintegration
and form new radioactive elements.
The disintegration is accompanied by
the emission of α, β or γ rays
The disintegration is at random i.e. and
every atom has equal chance each for
disintegration at any time.
(iii)
The number of atoms that disintegrate
per second is directly proportional to the
number of remaining unchaged
radioactive atoms present at any time.
The disintegration is independent of all
physical and chemical conditions like
temperature, pressure, chemical
combination etc.
The two laws of radioactive disintegration can
be summed up as below.
1. Group displacement law. The result of
α-β particle changes can be summed up in
the form of group displacement law "in an α
particle change the resulting element has an
atomic weight less by four units and atomic
number less by two units and atomic number
less by two units and it falls in a group of the
periodic table two columns to the left of the
original element, and in a β-particle change
the resulting element has some atomic
weight but its atomic number is increased by
one than its parent and hence it lies one
column to its right."
However, group displacement law, also
known as Rutherford Soddy rules, is now a
days stated as follows
(i) The total electric charge (atomic
number) or algebric sum of the charges
before the disintegration must be equal
to the total electric charge after the
disintegration.
(ii)
The sum of the mass number of the
initial particles must be equal to the
sum of the mass number of the final
particles.
Remember that a daughter nuclide (a
nucleus of a specific mass number is called
nuclide) differs from its parent nuclide not
(57)

only in its radioactive properties, but also in
other physical and chemical properties and
thus we can say that the radioactive process
involves the transmulation of the element. For
example,
(i) Emission in α particle
88Ra 226 → 86 Rn 222 + 2 He 4
Radium
Radon
(Group II) (Group 0)
(ii) Emission of a β particle.
82Pb 210 → 83 Bi 210 + 1 β 0
Lead
Bismuth
(Group IV) (Group V)
18. Points to remember
1. α-Decay produces isodiapher i.e. the
parent and daughter nuclide formed by
α-decay have same isotopic number, i.e.
difference between the number of neutrons
and protons is same. For example
88Ra 226 → 86 Rn 222
No, of neutrons 138 136
No. of protons 88 86
Difference 50 50
Thus note that an a decay leads to
(i)
(ii)
(iii)
decrease in atomic weight, mass
number and number of nucleons by four
units.
increse in number of protons, neutrons,
nuclear charge and atomic number by
two units.
increse in n/p ratio
2. β-Decay results in the formation of the
isobaric element i.e., parent and daughter
nuclide have different atomic numbers but
same mass number. For example
AISECT TUTORIALS : CHEMISTRY : SET-6
(58)
19K 40 → 20 Ca 40 + − 1e 0
Thus note that a β-decay leads to
(i)
(ii)
(iii)
no change in atomic weight, mass
number and number of nucleons.
decrease in number of neutrons by one
unit.
increase in nuclear charge number of
protons and atomic number by one unit.
It is important to note that although β-particle
(electron) is not present in the nucleus even
then it is emitted from the nucleus since a
neutron at first breaks down to a proton and
electron
0n 1 → 1 P 1 + −1e 0
The proton is retained by the nucleus when
the electron is emitted as a β-particle
3. Emission of 1 α−particle and 2 β−particles
in succession produces an isotope of the
parent element. For example,
92U 235
−α
Th 231 −β
→ 90
→ a1 Pa231
−β
→ 92 U 231
2. Law of radioactive decay. According to
the law of radioactive decay, the quantity of
a radioelement which disappers in unit time
(rate of disintegration) is directly proportional
to the amount present.
Determination of the number of a-and
b-particles emitted in a nuclear reaction.
Consider the following general reaction
m m
n X → 1 n 1 γ + a 4 2 α + b 0 −1 β
Then
(i) m=m' + 4 a + 0b
(ii) n=n' + 2 a-b
Solve for a and b
where a is the number of
4
2
He emitted

AISECT TUTORIALS : CHEMISTRY : SET-6
b is the number of
0 −1β
emitted
18. Points to remember.
1. Rate of decay is the number of atoms
undergoing decay in unit time, it is
represented by
- dN t
dt
2. Rate of decay of a nuclide is directly
proportional to the number of atoms of that
nuclide present at that moment, hence
− dN t
∝ N or dN t
= λ N t
dt dt
(The word d indicates a very-very small
fraction; the negative sign shows that the
number of radioactive atoms N t
decreases as
time t increses)
3. Rate of decay of nuclide is independent of
temperature so its energy of activation is
zero.
4. Since the rate of decay is directly
proportional to the amount of the radioactive
nuclide present and as the number of
undecomposed atoms decreases with
increase in time, the rate of decay also
decreases with the increase in time.
Various forms of equation for radioactive
deacy are
Nt = N 0 e −λt
log No-log N t
=0.4343 λt
log No =
λt
N t 2.303
λ = 2.303
t
log No
N t
Note that this equation is similar to that of first
order reaction, hence we can say that
radioactive disintegrations are examples of
first order reactions.
where
No = Initial number of atoms of the given
nuclide, i.e. at time 0
N t
λ
= No. of atoms of that nuclide present
after time t
= Decay constant
However, unlike first order rate constant (K)
the decay constant (λ) is independent of
temperature.
Decay constant. The ratio between the
number of atoms disintegrating in unit time to
the total number of atoms present at that time
is called the decay constant of that nuclide.
Characteristics of decay constant (λ)
1. It is characteristic of a nuclide (not for an
element)
2. Its units are time -1
3. Its value is always less than one.
Illustration 9. An element X with atomic
number 90 and mass number 232 loses one
α and two β particles successively to give a
stable speices Z. What would be the atomic
number and atomic weight of Z ?
(C.P.M.T.)
Solution At. No. and at. wt. of the element
(say Y) produced by the loss of one α-particle
( 2
He 4 ) from 90
X 232 = 88
Y 228
Illustration 10. Find out the total number of a
and β-particles emitted in the disintegration of
90 Th232 to 82
lead 208 (C.P.M.T.)
Solution
Change in at. wt. = 232-208 =24 awu (at.wt.
unit)
(59)

AISECT TUTORIALS : CHEMISTRY : SET-6
Now since in one α -particle emission at. wt.
is decreased by 4 awu, the number of
α-emissions for 24 awu = 24/4 =6
Atomic number after 6α emissions
=90-12=78 ( . . . a= 2
He 4 )
Increase in atomic number from 78 to the
given 82 = 82-78=4 ( . .. β particle = -1e 0 )
. . . No of β particle emissions = 4
Illustration 11 92
U 235 belongs to group III B
of the periodic table. It loses one α particle to
form the new element. Predict the position of
the new element in the periodic table.
(M.L.N.R.)
Solution. Since loss of an α-particle
decreases the atomic number of the element
by 2, the resulting product will lie two group to
the left of the parent group. Thus in the
present case the element will lie in I. A group
of the periodic table.
Illustration 12 92
U 238 lies in the VI B group of
the periodic table. It loses 8 α-particles and 6
β particles to form a new element. Predict the
position of the new element in the periodic
table.
Solution. Since loss of each α particle
displaces the daughter element to the left by
2 place, loss of 8 α-particles will displace the
element to the left by 16 places.
Further, loss of one β-particle displaces the
element one place to the right. The total
displacement towards right by the loss by 6
β-particles will be 6.
Thus the net displacement of the product wil
be 16-6 =10 places of the left. For knowing
the position after 10 displacement, we must
review the basic skeleton of the periodic table
IB IIB IIIB IVB VB VIB VIIB VIII
←⎯⎯⎯⎯⎯ ⎯
5 position displacement
towards left
VIII IA IIA IIIA IVA VA VIA VIIA o
←⎯⎯⎯⎯⎯⎯⎯⎯⎯
Next 5 displacement to left
Thus the new element will lie in IVA group of
the periodic table.
Half life period (T1/2 or t 1/2)
Rutherford in 1904 introduced a constant
known as half life period of the radioelement
for evaluating its radioactivity or for
comparing its radioactivity with the activities
of other radioelements. The half life period of
a radio-element is defined as the time
required by a given amount of the element to
decay to one half of its initial value.
Mathematically, T 1/2
= 0.693
λ
Now since λ is a constant we can conclude
that half life period of a particular
radioelement is independent of the amount of
the radioelement. In other words whatever
might be the amount of the radioactive
element present at a time, it will always
decompose to its half at the end of one half
life period. This can beautifully be expressed
in the form of a table where x represents the
amount of substance at start (i.e. when
time=o) and T 1/2
represents one half period of
the element.
Half life period is a measure of the
radioactivity of the element since shorter the
half life period of an element greater is the
number of the disintegrating atoms and
hence greater is its radioactivity. The half life
periods or the half lives of different
(60)

adioelements vary widely, ranging form a
fraction of a second to millions of years.
19. Average Life Period (T).
Since total decay period of any element is
infinity it is meaningless to use the term total
decay period (total lie period) for
radioelements. Thus the term average life is
used which is determined by the following
relation
Sum of lives of the nuclei
Average life(T) =
Total number of nuclei
Relation between average life and half life.
Average life(T) of an element is the
inverse of its decay constant i.e.
T = 1 λ
Substituting the value of λ in the above
equation,
T = T 1/2
0.693
Thus Average life (T) = 1.44 x Half life (T 1/2
)
= 2 x T 1/2
Thus the average life period of a radioisotope
is approximately under-root two time of its
half life period.
Spedific activity. It is the measure of
radioactivity of a rdioactive substance. It is
defined as the number of radioactive nuclei
which decay per second per gram of
radioactive isotope., Mathematically, If 'm' is
the mass of radioactive isotop then
Rate of decay
Specific activity =
= λ x
Avogadro number
Atomic mass in g
m = λN m
Where N is the number of radioactive nuclei
which undergoes disintegration
AISECT TUTORIALS : CHEMISTRY : SET-6
(61)
20. Determination of Number of Half
Life and Time of Decomposition
Half period of a radioactive substance is the
time when the concentration reduces to half
of its initial value which is represented by T 1/2
.
This value is constant for a given substance.
Suppose initially i.e. t=0 the concentration of
radioactive element =No and T=T 1/2
the
number rof undecomposed atoms =N1
. . .N i = N0
= N
2 0
⎛ 1 ⎞ ⎝ 2 ⎠
1
In the same manner t=2T 1/2
(Two half life) the
number of undecomposed atoms are N 2
N 2 = N1
= N0
= N
2 4 0
⎛ 1 ⎞ ⎝ 2 ⎠
and t=-3T1/2 (Three half life) the number of
radio element
undercomposed after three half life period is
N 3
N 3 = N2
2 = N0
8
= N 0 ( 1 2 )3
Similarly when t=nT 1/2
the number of
undercomposed radioacive atoms.
N n = N 0 ( 1 2 )n
In other words after n half life period the
number of residual atom would br No ( 1 .
2 )n
So if half life period of a radioactive element
is known, then time of decomposition can be
calculated as follows
Time of decomposition= Half life x No. of half
life.
Illustration 13. Calculate the half life period
of an isotope if its disintegration constant is
1.237 x 10 -4 year -1 .
Solution We know that
T 1/2 = 0.693
λ
. . . T 1/2 =
0.693
= 5600 years
−4
1.237 x 10
2

AISECT TUTORIALS : CHEMISTRY : SET-6
Illustration 14. Calculate the disintegration
constant and mean life of the radioactive
isotope of hydrogen whose half life is 12.25
years.
Solution . We know that
λ = 0.693
t 1/2
= 0.693
12.25
= 5.652 x 10 -2 year -1
Again we know that
Mear life = 1 λ
=
1
5.652 x 10 −2 = 17.7 years
Illustration 15. The half life period of radium
is 1620 years. In how much time 1 gm of
radium will reduce to 0.25 gm ?
Solution 1 gm of Ra wil reduce to 0.5 gm in
1620 yrs.
0.5 gm of Ra wil reduce to 0.25 gm in another
1620 yrs.
. . . Time taken by 1 gm of Ra to reduce to
0.25 gm = 1620 + 1620 = 3240
Illustration 16. Radioactivity is a first order
process. Radioactive carbon in wood sample
decays with a half life of 5770 years. What
fraction would remain after 11540 years ?
Solution. We know that after 5770 years
(T 1/2
) half of the original amount is left.
Therefore after another 5770 years (i.e. after
5770+ 5770= 11540 years of the start) half of
the remaining half will decompose and thus
the fraction left will be
= 1 − ( 1 2 + 1 2 x 1 2 )
= 1 − ( 1 2 + 1 4 )
= 1 4
Thus 25% of the original sample will be
left after 11540 years.
Illustration 17 The half life of radon is 3.80
days. Calculate the time in which its one
twentieth of the left behind.
Solution. Since λ = 0.693
3.80
= 0.182 per day
per day
0.693
T 1/2
Now let the initial amount (No) be a then N t
=
a/20.
. . . t= 2.303
0.182
= 2.303
0.182
log
NO
NT = 2.303
0.182 log a
a/20
log 20 = 16.45 days.
21. Radioactivity is a Nuclear Property
(i) The new elements are formed from the
emission of α or β particles.
(ii) Radioactivity does not depend upon the
state of radioactive sample. The activity is the
same whether it is a metal or its hydroxide or
any other compound.
(iii) Radioactive changes are spontaneous.
These are not' controlled by temperature.
pressure or nature of chemical combination.
Kinetically it is a first order reaction.
(iv) Radioactive phenomenon or rate of decay
is not affected when the radioactive sample is
placed in electric, magnetic or gravitational
fields.
21.1 Measurement of Radioacivity
A series of instruments have been designed
for determination of radioactivity. These are
as follows
(i)
(ii)
(iii)
(iv)
(v)
Electroscope;
Wilson cloud chamber;
Bubble chamber method;
Geiger-Muller counter; and
Scintillation counter based on use of
zinc sulphide screen.
(62)

AISECT TUTORIALS : CHEMISTRY : SET-6
21.2 Units of Radioactivity.
The standard unit in radioactivity is curie (c)
which is defined as that amount of any
radioactive material which gives 3.7 x 10 10
disintegrations per second (dps), i.e.
IC = 3.7 x 10 10 dps.
The millicurie (mc) and microcurie ((µc)
are
equal to 10 -3 and 10 -6 curies i.e. 3.7x 10 7 and
3.7 x 10 4 dps respectively.
In short
1c = 10 3 mc=10 6 µ c
1c = 3.7 x 10 10 dps
1mc = 3.7=10 7 dps
1µc = 3.7=10 4 dps
But now a days, the unit curie is replaced by
rutherford (rd) which is defined as the
amount of a radioactive substance which
undergoes 10 6 dps. i.e. 1 rd = 10 6 dps. The
millcurie and microcurie corresponding
rutherford units are millirutherford (mrd) and
microrutherford (µrd) respectively.
1 c = 3.7 x 10 10 dps = 37 x 10 3 rd
1 mc = 3.7 x 10 7 dps = 37 rd
1 µc=3.7 x 10 4 dps =37 mrd
However in SI system the unit of radioactivity
is becquerel (Bq)
1 Bq= 1 disintegration per second
= 1 µrd
10 6 Bq = 1rd
3.7 x 10 10 Bq = 1c
22. Disintegration Series.
The phenomenon of natural radioactivity
continues till stable nuclei is formed. All the
nuclei from the initial element to the final
stable element constitute a series known as
disintegration series.Further we know that
mass numbers change only when α-particles
are emitted (and not when β particles are
emitted) causing the change in mass of 4
units in each step. Hence the mass numbers
of all elements in a series will fit into one of
the following formulae.
4n, 4n +1, 4n+2 or 4 n+3
Hence there can be only four disintegration
series (see table)
The numbers indicate that in a particular
series the mass numbers of all the members
are either divisible by 4 (in case of 4n) or
divisible by 4 with remainder of 1,2 or 3 (in
the rest three series) n being an integer. In
other words, the mass numbers of the
members of 4n, 4n + 1 , 4n + 2 and 4n + 3
series are exactly divisible by 4, 4 + 1, 4 + 2
and 4 + 3 respectively.
Points to remember. 1. 4n + 1 series is an
artifical series while the rest three are natural.
2. The end product in the 4n + 1 series is
bismuth, while in the rest three a stable
isotope of lead is the end product.
3. The 4n + 1 series starts from plutonium
94 Pu241 but commonly known as
nepotunium series because neptunium
is the longest lived member of the
series.
Series 4n 4n +1 4n +2 4n+3
n 58 59 59 58
Parent
element
Half life
Prominant
element
90 Th 232 94 Th 241 92 U 238 92 U 235
1.39 x
10 10
years
10 years 4.5 x 10 9
years
7.07 x 10 8
years
Th 232 NP 237 U 238 227
90 93 92 89AC (63)

AISECT TUTORIALS : CHEMISTRY : SET-6
Half life
Name of
Series
End
product
1.39 x
10 10
years
Thorium
(Natural)
2.2 x 10 6
years
Naptunium
(Artifical)
4.5 x 10 9
years
Uranium
(Natural)
13.5years
Actinium
(Natural)
Pb 208 Bt 209 Pb 206 207
82 83 92 82Pb n 52 52 51 51
Number of
lost
particles
α=6
β=4
α=8
β=5
α=8
β=6
α=7
β=4
4. The 4n +3 series actually starts from 92
U 235
The total number of α and β particles emitted
in a series can be easily obtained from a
knowledge of the first and last members of
the series as illustrated in illustration 10.
Illustration 18. To which disintegration sereis
do the Ra -226 and U-235 belong ?
Solution Divide the mass number of the
element by four and find out the remainder. In
case of Ra-226, it is 2 while in U-235 it is 3.
Hence Ra-226 belongs to 4n + 2 series and
U-235 belongs to 4n + 3 series.
23. Radioactive Equlibrium.
Suppose a radioactive element A
disintegrates to form another radioactive
element B which in turn disintegrates to still
another element C.
A → B → C
In the begining, the amount of A (in terms of
atoms) is large while that of B is very small.
Hence the rate of disintegration of A into B is
high while that of B into C is low. With the
passage of time. A goes on disintegrating
while more and more of B is formed. As a
result, the rate of disintegration of A to B goes
on decreasing while that of B to C goes on
increasing. Ultimately, a stage is reached
when the rate of disintegration of A to B is
equal to that of B to C with the result the
amount of B remains constant. Under these
conditions B is said to be in equilibrium with
A. For a radioactive equilibrium to be
established half life of the parent must be
much more than half life of the daughter.
It is important to note that the term
equilibrium is used for reversible reactions
but the radioactive reactions are irreversible,
hence it is prefrered to say that B is in a
steady state rather than in equilibrium state.
Thus at a steady state,
Mathematically,
NA
= λB = TA (.
NB λA TB . . T = 1 ) λ
Where λa and λb are the radioactive
constants for the processes A B and BC
respectively.
where T A
and T B
are the average life periods
of A and B respectively.
In terms of half life periods
NA
= (T 1/2) A
NB (T 1/2 ) B
Thus at a steady state (at radioactive
equilibrium), the amounts (number of atoms)
of the different radioelements present in the
reaction series are inversely proportional to
their radioactive constants or directly
proportional to their half life and also average
life periods.
It is important to note that the radioactive
equilibrium difers from ordinary chemical
equilibrium because in the former the amount
of the different substances involved are not
constant and the changes are not reversible.
(64)