MOTION SAMPLE BOOKLET [CLASS XII]
• GRAVITATION
• CHEMICAL KINETICS
• FUNCTION
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THEORY AND EXERCISE BOOKLET
CONTENTS
GRAVITATION
S.NO.
TOPIC
PAGE NO.
THEORY WITH SOLVED EXAMPLES ...................................................
5 – 29
EXERCISE - I .........................................................................................
30 – 37
EXERCISE - II ........................................................................................
38 – 44
EXERCISE - III ........................................................................................
45 – 49
EXERCISE -IV ........................................................................................
50 - 58
ANSWER KEY .......................................................................................
59 - 60
CHEMICAL KINETICS
S.NO.
TOPIC
PAGE NO.
THEORY WITH SOLVED EXAMPLES ...................................................
61 – 101
EXERCISE - I .........................................................................................
109 – 116
EXERCISE - II ........................................................................................
117 – 129
EXERCISE - III ........................................................................................
130 – 141
EXERCISE - IV .......................................................................................
142 – 151
ANSWER KEY .......................................................................................
152 – 155
FUNCTION
S.NO.
TOPIC
PAGE NO.
THEORY WITH SOLVED EXAMPLES ...................................................
156 – 179
EXERCISE - I .........................................................................................
180 – 186
EXERCISE - II ........................................................................................
187 – 193
EXERCISE - III ........................................................................................
194 – 212
EXERCISE - IV .......................................................................................
213 – 218
ANSWER KEY .......................................................................................
219 - 222
SYLLABUS
•
GRAVITATION
Gravitational potential and field; Acceleration due to gravity; Motion of planets and
satellites in circular orbits.
•
.
CHEMICAL KINETICS
Rates of chemical reactions; Order of reactions; Rate constant; First orderreactions;
Temperature dependence of rate constant (Arrhenius equation).
•
FUNCTION
Real valued functions of a real variable, into, onto and one-to-one functions, sum,
difference, product and quotient of two functions, composite functions, absolute value,
polynomial, rational, trigonometric, exponential and logarithmic functions,
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GRAVITATION
Page # 5
GRAVITATION
INTRODUCTION
Newton observed that an object, an apple, when released near the earth surface is accelerated
towards the earth. As acceleration is caused by an unbalanced force, there must be a force pulling
objects towards the earth. If someone throws a projectile with some initial velocity, then instead of
that object moving off into space in a straight line, it is continously acted on by a force pulling it back
to earth. If we throw the projectile with greater velocity then the path of projectile would be different
as well and its range is also increased with initial velocity. If the projection velocity is further increased
until at some initial velocity, the body would not hit the earth at all but would go right around it in an
orbit. But at any point along its path the projectile would still have a force acting on it pulling it toward
the surface of earth.
Newton was led to the conclusion that the same force that causes the apple to fall to the earth
also causes the moon to be pulled to the earth. Thus the moon moves in its orbit about the earth
because it is pulled toward the earth. But if there is a force between the moon and the earth, why not
a force between the sun and the earth or why not a force between the sun and the other planets ?
Newton proposed that the same force, named gravitational force which acts on objects near the earth
surface also acts on all the heavenly bodies. He proposed that there was a force of gravitation
between each and every mass in the universe.
1.1
Newtons's Law of Universal Gravitation
All physical bodies are subjected to the action of the forces of mutual gravitational attraction. The
basic law describing the gravitational forces was stated by Sir Issac Newton and it is called Newton's
Law. of Universal gravitation.
The law is stated as : "Between any two particles of masses mI and m2 at separation r from each other
there exist attractive forces FAB and FBA directed from one body to the other and equal in magnitude
which is directly proportional to the product of the masses of the bodies and inversely proportional to
the square of the distance between the two". Thus we can write
FAB
FBA
G
m1m2
...(1)
r2
Where G is called universal gravitational constant. Its value is equal to 6.67 × 10–11 Nm2/kg. The law of
gravitation can be applied to the bodies whose dimensions are small as compared to the separation
between the two or when bodies can be treated as point particles.
m1
FAB
FBA
m2
A
B
r
If the bodies are not very small sized, we can not directly apply the expression in equation-(1) to find
their natural gravitational attraction. In this case we use the following procedure to find the same. The
bodies are initially split into small parts or a large number of point masses. Now using equation-(1) the
force of attraction exerted on a particle of one body by a particle of another body can be obtained.
Now we add all forces vectorially which are exerted by all independent particles of second body on the
particle of first body. Finally the resultants of these forces is summed over all particles of the first body
to obtain the net force experinced by the bodies. In general we use integration or basic summation of
these forces.
Gravitational force is a conservative force.
Gravitational force is a central force.
Gravitational force is equal in magnitude & opposite in direction
Gravitational forces are action - reaction pair.
Gravitational force acts along the line joining the two masses.
Gravitational force doesn't depend upon the medium
Gravitational force is an attractive force.
–Gm1m2r
F
| r |3
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2.
2.1
[Head of r is placed at that position where we have to evaluate force]
GRAVITATIONAL FIELD
We can state by Newton's universal law of gravitation that every mass M produces, in the region
around it, a physical situation in which, whenever any other mass is placed, force acts on it, is called
gravitational field. This field is recognized by the force that the mass M exerts another mass, such as
m, brought into the region.
Strength of Gravitational Field
We define gravitational field strength at any point in space to be the gravitational force per unit mass
on a test mass (mass brought into the field for experimental observation). Thus for a point in space if
a test mass m0, experiences a force F , then at that point in space, gravitational field strength which
F
m0
is denoted by g , is given as g
Gravitational field strength g is a vector quantity and has same direction as that of the force on the
test mass in field.
Generally magnitude of test mass is very small such that its gravitational field does not modify
the field that is being measured. It should be also noted that gravitational field strength is just the
acceleration that a unit mass would experience at that point in space.
2.2
Gravitational Field Strength of Point Mass
As per our previous discussion we can state that every point mass also
produces a gravitational field in its surrounding. To find the gravitational
field strength due to a point mass, we put a test mass m0 at a point P at
distance x from a point mass m then force on m0 is given as
Gmm0
x2
Fg
Fg
m0
x
m
Now if at point P, gravitational field strength due to m is gp then it is given as
Fg
Gm
gp
m0
x2
The expression written in above equation gives the gravitational field strength at a point due to a point
mass.
It should be noted that the expression in equation written above is only applicable for gravitational
field strength due to point masses. It should not be used for extended bodies.
However, the expression for the gravitational field strength produced by extended masses has already
been derived in electrostatics section.
[Just replace k by –G & Q by M in those expression]
So we will just revise the expression of gravitational field strength at points due to various extended
masses. Gravitational field strength :
–GMx
(x 2 R 2 )3 /2
1.
At a point on the axis of Ring =
2.
2GM
x
At a point on the axis of disc = R2 1 –
2
R
x2
3.
At an axial point of a rod =
4.
Due to a circular Arc =
5.
Due to a long infinite thread =
6.
Due to long solid cylinder
(a) at an outer point =
–GM 1
1
–
L
x x L
–2GM sin( / 2)
R2
–2G
x
–2G R 2
x
(where
is mass density per volume)
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GRAVITATION
7.
Page # 7
(b) at an inner point = – 2G
Due to hollow sphere :
(a) for outer points =
–GM
x
2
(b) for points on surface =
x
–GM
(Behaving as a point mass)
R2
3.
x
–GM
(Behaving as a point mass)
x2
–GM
(b) For points on surface =
(c) For inner points =
1
x2
R2
Due to solid sphere
(a) For outer points =
g
–GM
(c) for inner points = 0 (As no mass is enclosed within it)
8.
R x
g
(Behaving as a point mass)
g
(Behaving as a point mass)
R2
–GM
–GMx
R
g
x
g
1
x2
R2
3
INTERACTION ENERGY
This energy exists in a system of particles due to the interaction forces between the particles of
system. Analytically this term is defined as the work done against the interaction of system forces in
assembling the given configuration of particles. To understand this we take a simple example of
interaction energy of two points masses.
Figure (a) shows a syst em of two point masses m1 and m2 placed at a distance r apart in space.
here if we wish to find the interaction potential energy of the two masses, this must be the work done
in bringing the two masses from infinity (zero interaction state) to this configuration. For this we first
fix m1 at its position and bring m2 slowly from infinity to its location. If in the process m2 is at a distance
x from m1 then force on it is
F
m1
–
Gm1m2 ˆ
i
x2
m1
m2
r
m2
r
Fdx
(a)
(b)
This force is applied by the gravitational field of m1 to m2. If it is further displaced by a distance dx
towards m1 then work done by the field is
dW
F . dx =
Gm1m2
x2
dx
Now in bringing m2 from infinity to a position at a distance r from m1, the total work done by the field is
r
W
dW –
W
Gm1m2
1
dx = – G m m –
1
2
x2
x
r
Gm1m2
r
Gm1m2
amount of work. The work is positive
r
because the displacment of body is in the direction of force.
Initially when the separation between m1 and m2 was very large (at infinity) there was no
interaction between them. We conversely say that as a reference when there is no interaction the
interaction energy of the system is zero and during the process system forces (gravitational forces)
are doing work so system energy will decrease and becomes negative (as initial energy was zero). As
a consequence we can state that in general if system forces are attractive, in assembling a system of
particles work will be done by the system and it will spend energy in assembling itself. Thus finally the
interaction energy of system will be negative. On the other hand if in a given system of particles, the
Thus during the process field of system has done
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system forces are repulsive, then in assembling a system some external forces have to be work against
the system forces and in this case some work must be done by external forces on the system hence
finally the interaction energy of the system of particles must be positive.
In above example as work is done by the gravitaional forces of the system of two masses, the
interaction energy of system must be negative and it can be given as
Gm1m2
U12 –
...(1)
r
As gravitational forces are always attractive, the gravitational potential energy is always taken negative.
3.1
Interaction Energy of a System of Particles
If in a system there are more than two particles then we can find the interaction energy of particle in
pairs using equation (1) and finally sum up all the results to get the total energy of the system. For
example in a system of N particles with masses m1, m2........mn separated from each other by a
distance r12 ........ where r12 is the separation between m1 and m2 and so on.
In the above case the total interaction energy of system is given as
1 N N Gmm
i
j
U –
2 i 1 j 1 rij
1
is taken because the interaction energy for each possible pair of
2
system is taken twice during summation as for mass m1 and m3
In this expression the factor
U
Ex.1
Sol.
–
Gm1m3
Gm3m1
= –
r13
r31
Now to understand the applications of interaction energy we take few examples.
Cm
Three particles each of the mass m are placed at the corners
of an equilateral triangle of side d and shown in figure.
60°
Calculate (a) the potential energy of the system, (b) work
d
done on this system if the side of the triangle is changed
from d to 2d.
(a) As in case of two-particle system potential energy is
A
given by (–Gm1m2/r), so
m
2
Gmm
3Gm
U
–3
–
U = U12 + U23 + U31 or
i
d
d
(b) When d is changed to 2d,
B
m
3Gm2
2d
Thus work done in changing in potential energy is given as
Uf = –
W = U f – Ui =
Ex.2
Sol.
3Gm2
2d
Two particles m1 and m2 are initially at rest at infinite distance. Find their relative velocity of
approach due to gravitational atraction when their separation is d.
Initiallly when the separation was large there was no interaction energy and when they get closer the
system gravitational energy decreases and the kinetic energy increases.
When separation between the two particles is d, then according to energy conservation we have
1
m1v12
2
Gm1m2
1
m2v 22 –
2
d
0
As no other force is present we have according to momentum conservation
m1v1 = m2v2
From equations written above
1
m1v12
2
1 m21 2
v1
2 m2
Gm1m2
d
or
v1
2Gm22
d(m1 m2 )
2G
m2
d(m1 m2 )
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And on further solving we get
v2
2G
m1
d(m1 m2 )
Thus approach velocity is given as
vap
Ex.3
Sol.
v1
v2
2G(m1 m2 )
d
If a particle of mass 'm' is projected from a surface of bigger sphere of mass '16M' and radius
'2a' then find out the minimum velocity of the paticle such that the particle reaches the surface
of the smaller sphere of mass M and radius 'a'. Given that the distance between the centres of
two spheres is 10 a.
When the particle is at the surface of bigger sphere it is
16M
M
attracted more by the bigger sphere and less by the smaller
sphere. As it is projected the force of attraction from bigger
2a
x
a
m
sphere decreases and that from smaller sphere increass and
thus the particle reaches the state of equilibrium at distance x
10a
from the centre smaller sphere
GMm
x2
G(16M)m
(10a – x)2
M
16M
m
(10a – x)2 = 16x2
2a
8a
10a – x = 4x
x = 2a
After this point the attraction on the particle from the smaller sphere becomes more than that from the
bigger sphere and the particle will automatically move towards the smaller sphere. Hence the minimum
velocity to reach the smaller sphere is the veloicty required to reach the equilibrium state according to
energy conservation, we have,
–
G(16M)m GMm
–
2a
8a
v2
4
1
mv2
2
–G(16M)m GMm
–
8a
2a
45GM
4a
v
45GM
4a
GRAVITATIONAL POTENTIAL
The gravitational potential at a point in gravitational field is the gravitional potenial energy per unit
mass placed at that point in gravitational field. Thus at a certain point in gravitational field, a mass m0
has a potential energy U then the gravitational potential at that point is given as
U
m0
or if at a point in gravitational field gravitational potential V is known then the interaction potential
energy of a point mass m0 at that point in the field is given as
U = m0v
Interaction energy of a point mass m0 in a field is defined as work done in bringing that mass from
infinity to that point. In the same fashion we can define gravitational potential at a point in field,
alternatively as "Work done in bringing a unit mass from infinity to that point against gravitational
forces."
V=
When a unit mass is brought to a point in a gravitational field, force on the unit mass is g at a point in
the field. Thus the work done in bringing this unit mass from infinity to a point P in gravitational field or
gravitational potential at point P is given as
P
VP
– g .dx
Here negative sign shown that VP is the negative of work done by gravitation field or it is the external
required work for the purpose against gravitational forces.
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4.1
Gravitational Potential due to a Point Mass
We know that in the surrounding of a point mass it produces its gravitational field. If we wish to find
the gravitational potential at a point P situated at a distance r from it as shown in figure, we place a
t est mass m0 at P and we find the interaction energy of m0 with the field of m, which is given as
P
Gmm0
U –
r
r
Now the gravitational potential at P due to m can be written as
V
U
m0
Gm
r
–
m
The expression of gravitational potential in equation is a standard result due to a point mass which can
be used as an elemental form to find other complex results, we'll see later.
The same thing can also be obtained by using equation
P
VP
r
g. dx
4.2
Gravitational potential
1.
Due to a rod at an a xial point = – G ln
2.
Due to ring at an axial point =
3.
Due to ring at the centre =
4.
Due to Disc = –G 2 [ R2
Due to hollow sphere
5.
for outer points =
VP
or
a
R2
a
(where
x2 – x]
is mass density per unit area)
R
Vg
–GM
R
GM
R
Vg
r=R
r
–GM
r
For surface points =
For inner points =
l
–GM
R
Due to solid sphere
For outer points =
Gm
r
x2
–GM
For surface points =
R
6.
–
–GM
–GM
r
For inner points = –
Gm
dx or VP
x2
–3GM
2R
–GM
R
–GM
(3R 2 – r 2 )
2R 3
Potential energy of hollow sphere =
Potential energy of solid sphere =
–GM2
2R
–3GM2
5R
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GRAVITATION
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Page # 11
GRAVITATIONAL LINES OF FORCES
Gravitational field can also be represented by lines of force. A
line of force is drawn in such a way that at each point the
direction of field is tangent to line that passes through the
point. Thus tangent to any point on a line of force gives the
direction of gravitational field at that point. By convention
lines of force are drawn in such a way that their density is
proportional to the strength of field. Figure shown shows the
field of a point mass in its surrounding. We can see that the
lines of force are radially inward giving direction of field and as
we go closer to the mass the density of lines is more which
shows that field strength is increasing.
Figure shown shows the configuration of field lines for a system
of two equal masses separated by a given distance.
Here we can see that there is no point where any two lines of
force intersects or meet. The reason is obvious that at one
point in space there can never be two direction of gravitational
fields. It should be noted that a line of force gives the direction
of net gravitational field in the region. Like electric field
gravitational field never exists in closed loops.
•
Gravitational Flux :
•
Gravitational Gauss law :
5.1
•
5.2
5.3
g.ds
g.ds
–4 GMin
Here g is the gravitational field due to all the masses. Min is the mass inside the assumed gaussian
surface.
Gravitational Field Strength of Earth:
We can consider earth to be a very large sphere of mass Me and radius Re. Gravitational field strength
due to earth is also regarded as acceleration due to gravity or gravitational acceleration. Now we find
values of g at different points due to earth.
A
Earth behaves as a non conducting solid sphere
Me
gs
Value of g on Earth's Surface :
If gs be the gravitational field strength at a point A on the
Re
surface of earth, then it can be easily obtained by using
Earth
the result of a solid sphere. Thus for earth, value of gs can
be given as
GMe
P
gs
...(1)
R2e
g h
Value of g at a Height h Above the Earth's Surface:
Me
If we wish to find the value of g at a point P as shown in figure
shown at a height h above the Eath's surface. Then the value can
be obtained as
Re
GMe
gs
gs
2
2
GMe
h
h
gs
2
or
2
R
1
1
e
(R e h)
R
R
e
e
If point P is very close to Earth's surface then for h << Re we can rewrite the expression in given
equation as
gh
gs 1
h
Re
–2
~
–
gs 1 –
2h
Re
[Using binomial approximation]
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5.4
Value of g at a Depth h Below the Earth's Surface
If we find the value of g inside the volume of earth at a depth h below the earth's surface at point P as
shown in figure, then we can use the result of g inside a solid sphere as
Me
h
GMe x
gin
R 3e
Re
R e–h
Here x, the distance of point from centre of earth is given as
x = Re – h
GMe(R e – h)
Thus we have gh
R 3e
gs 1 –
h
Re
...(3)
From equation (1), (2) and (3) we can say that the value of g at eath's surface is maximum and
as we move above the earth's surface or we go below the surface of earth, the value of g decrease.
5.5
geff
Effect of Earth's Rotation on Value of g
Let us consider a body of mass m placed on Earth's
surface at a latitude as shown in figure. This mass
experiences a force mgs towards the centre of earth and
a cent rifugal force mwe2 Re sin relative to Earth's surface
as shown in figure.
If we consider geff as the effective value of g on earth
surface at a latitude then we can write
F
= net = geff
m
4
e
( 2eRe sin )2
g2
2 e2R e sin .g cos(90
e
pole
R e sin
m
m
2
Re sin
mg
Re
) equator
is very very small
So we can write
geff
geff
g 1
g2
2 e2R e sin2 g
2 2eR e sin2
g
1/2
g–
2
e
R sin2
...(i)
From equation (1) we can find the value of effective gravity at poles and equatorial points on Earth as
At poles = 0
gpoles = gs = 9.83 m/s2
At equator
gequator = gs – 2Re = 9.78 m/s2
2
Thus we can see that the body if placed at poles of Earth, it will only have a spin, not circular motion
so there is no reduction in value of g at poles due to rotation of earth. Thus at poles value of g on
Earth surface is maximum and at equator it is minimum. But an average we take 9.8 m/s2, the value of
g everywhere on earth's surface.
5.6
=
Effect of Shape of Earth on Value of g
Till now we considered that earth is spherical in its shape but this is not actually true. Due to some
geological and astromonical reasons, the shape of earth is not exact spherical. It is ellipsoidal.
As we've discussed that the value of g at a point on earth surface depends on radius of
Earth.It is observed that the approximate difference in earth's radius at different points on equator and
poles is re – rp
~
–
21 to 34 km. Due to this the difference in value of g at poles and equatorial points is
approximately gp – ge
~
–
0.02 to 0.04 m/s2, which is very small. So for numerical calculations, generally,,
we ignore this factor while taking the value of g and we assume Earth is spherical in shape.
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GRAVITATION
Ex.4
Sol.
Calculate the mass and density of the earth. Given that Gravitational constant G = 6.67 × 10–11
Nm2/kg2, the radius of the earth = 6.37 × 106 m and g = 9.8 m/s2.
The acceleration due to gravity on earth surface is given as
ge =
If
Sol.
GMe
R2e
gsR e2
G
Me
or
9.8 (6.37 106 )2
6.67 10–11
4
R3
3
or
=
3M
4 R
3
=
Sol.
2
Re
1
0.99
2
or
g = 1.02 g
2
2
Re
gs = 2 × 3.14
6.4 106
9.8
5074.77 s
~
–
84.57 min.
Calculate the acceleration due to gravity at the surface of Mars if its diameter is 6760 km and
mass is one-tenth that of earth. The diameter of earth is 12742 km and acceleration due to
gravity on earth is 9.8 m/s2.
g
We know that
So
Sol.
g
gs
Re
or
Thus length of the day will be T
Ex.8
g'
At what rate should the earth rotate so that the apparent g at the equator becomes zero?
What will be the length of the day in this situation ?
At earth's equator effective value of gravity is
geq = gs – 2R e
If geff at equator is zero, we have
gs =
Ex.7
3 (6 1024 )
= 5.5 × 103 kg/m3
4 3.14 (6.37 106 )3
If the radius of the earth were to shrink by one percent, its mass remaining the same, what
would happen to the acceleration due to gravity on the earth's surface?
Consider the case of body of mass m placed on the earth's surface (mass of the earth M and radius R).
If g is acceleration due to gravity, then we known that
GMe
gs =
...(1)
R2e
Now, when the radius is reduced by 1%, i.e. radius becomes 0.99 R, let acceleration due to gravity be
g', then
GM
g'
...(2)
(0.99 R)2
From equation (1) and (2), we get
g'
R2
1
or
g (0.99R) (0.99)2
Thus, the value of g is increased by 2%.
Ex.6
= 6 × 1024 kg
be the density of earth, then
M=
Ex.5
Page # 13
gM
gE
MM
ME
RE
RM
2
GM
R2
1
10
12742
6760
2
gM
gE
0.35 or g = 9.8 × 0.35 = 3.48 m/s2
M
Calculate the apparent weight of a body of mass m at a latitude when it is moving with speed
v on the surface of the earth from west to east at the same latitude.
If W be the apparent weight of body at a latitude
then from figure shown, we have
W = mg – m 2R cos2
...(1)
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Re cos
When body moves at speed v from west to east
relative to earth, its net angular speed can be
given as
v
[ e earth's angular velocity]
e
Rcos
Now from equation (1) we have
W
mg – m
or W
e
mg – m
mg – m
2
e
R cos 2
w2R cos2
mg 1 – e
g
v2
R cos2
–
mv 2
R
2
R e cos
Re
2 ev
R cos
– 2m
2
mgs
R cos2
2
–
m
2
v
R cos
2
e
N
m
R e cos
e
equator
R cos2
v cos
e
e
v cos
g
mv2
[Neglecting
as being very small]
R
6.
SATELLITE AND PLANETARY MOTION
6.1
Motion of a Satellite in a Circular Orbit
To understand how a satellite continously moves in its orbit, we consider the projection of a body
horizontally from the top of a high mountain on earth as shown in figure. Here till our discussion ends
we neglect air friction. The distance the projectile travels before hitting the ground depends on the
launching speed. The greater the speed, the greater the distance. The distance the projectile travels
before hitting the ground is also affected by the curvature of earth as shown in figure shown. This
figure was given by newton in his explanantion of laws of gravitation. it shows different trajectories for
diferent launching speeds. As the launching speed is made greater, a speed is reached where by the
projectile's path follow the curvature of the earth. This is the launching speed which places the
projectile in a circular orbit. Thus an object in circular orbit may be regarded as falling, but as it falls its
path is concentric with the earth's spherical surface and the object maintains a fixed distance from the
earth's centre. Since the motion may continue indefinitely, we may say that the orbit is stable.
v
Earth
Fe
Let's find the speed of a satellite of mass m in a circular
orbit around the earth. Consider a satellite revolving around
the earth in a circular orbit of radius r as shown in figure.
If its orbit is stable during its motion, the net gravitational
force on it must be balanced by the centrifugal force on it
relative to the rotating frame as
GMem
r2
6.2
mv2
or
r
v
m
FG
v
r
Me
Re
GMe
r
Earth
Expression in above equation gives the speed of a statellite
in a stable circular orbit of radius r.
Energies of a Satellite in a Circular Orbit
When there is a satellite revolving in a stable circular orbit of radius r around the earth, its speed is
given by above equation. During its motion the kinetic energy of the satellite can be given as
K
1
mv2
2
1 GMem
2 r
...(1)
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As gravitational force on satellite due to earth is the only force it experiences during motion, it has
gravitational interaction energy in the field of earth, which is given as
GMem
...(2)
r
Thus the total energy of a satellite in an orbit of radius r can be given as
Total energy E = Kinetic energy K + Potential Energy U
U –
1 GMem GMem
–
2 r
r
E
or
–
1 GMem
2 r
From equation (1), (2) and (3) we can see that | k |
...(3)
U
|E|
2
The above relation in magnitude of total, kinetic and potential energies of a satelline is very useful in
numerical problem so it is advised to keep this relation in mind while handing satellite problems related
to energy.
Now to understand satellite and planetary motion in detail, we take few example.
Ex.9
Sol.
Estimate the mass of the sun, assuming the orbit of the earth round the sun to be a circle. The
distance between the sun and earth is 1.49 × 1011 m and G = 6.66 × 10–11 Nm2/kg 2.
Here the revolving speed of earth can be given as
GM
r
v
[Orbital speed]
Where M is the mass of sun and r is the orbit radius of earth.
We known time period of earth around sun is T = 365 days, thus we have
T=
2 r
v
or T
2 r
4 (3.14)2 (1.49 1011 )3
r
4 2r3
or M =
=
= 1.972 × 1022 kg
(365 24 3600)2 (6.66 10–11 )
GM
GT2
Ex.10 If the earth be one-half of its present distance from the sun, how many days will be in one year ?
Sol.
If orbit of earth's radius is R, in previous example we've discussed that time period is given as
T
2 r
r
Gm
2
GM
r3 / 2
r
, new time period becomes
2
2
T'
r '3 / 2
GM
From above equations, we have
If radius changes or r =
T
T'
r
r'
3/2
or
T'
T
r'
r
3/2
= 365
1
2
3/2
365
2 2
days
Ex.11 A satellite revolving in a circular equatorial orbit of radius r = 2.00 × 104 km from west to east
appear over a certain point at the equator every t = 11.6 hours. Using this data, calculate the
mass of the earth. The gravitational constant is supposed to be known.
Sol.
Here the absolute angular velocity of satellite is given by
= s+ E
Where E is the angular velocity of earth, which is from west to east.
or
2
t
2
T
From Kepler's III law, we have
[Where t = 11.6 hr. and T = 24 hr.]
GM
r3 /2
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Thus we have
M
or
GM
r3 / 2
2
t
4 2r 3 1
G
t
1
T
2
T
2
4 2 (2 107 )3
1
–11
(6.67 10 ) 11.6 3600
1
24 3600
2
= 6.0 × 1024 kg
Ex.12 A satellite of mass m is moving in a circular orbit of radius r. Calculate its angular momentum
with respect to the centre of the orbit in terms of the mass of the earth.
Sol.
The situation is shown in figure
The angular momentum of the satellite with respect to the centre of orbit is given by
L
Where r
centre
r
mv
v
is the position vector of satellite with respect to the
Satellite
.m
of orbit and v is its velocity vector of satellite.
In case of circular orbit, the angle between r and v is 90°. Hence
L = m v r sin 90° = m v r
...(1)
The direction is perpendicular to the plane of the orbit.
We know orbital speed of satellite is
GM
r
From equation (1) and (2), we get
v
r
M
Earth
...(2)
GM
L (GM m2 r)1 /2
r
Now we will understand the concept of double star system through an example.
L
m
Ex.13 In a double star, two stars of masses m1 and m2. distance d apart revolve about their common
centre of mass under the influence of their mutual gravitational attraction. Find an expression
for the period T in terms of masses m1, m2 and d. Find the ratio of their angular momenta about
centre of mass and also the ratio of their kinetic energies.
Sol.
The centre of mass of double star from mass m1 is given by
rcm
m1r1
m1
m2r2
m2
m1 0 m2d
m1 m2
m2 d
m1 m2
Distance of centre of mass from m2 is
r 'cm
d – rcm
d–
m2d
m1 m2
m1d
m1 m2
Both the stars rotate around centre of mass in their own
circular orbits with the same angular speed . the
gravitational force acting on each star provides the
necessary centripetal force. if we consider the rotation of
mass m1, then
m1(rcm)
2
This gives
Gm1m2
d2
m1
or
=
2
=
T
m2 d
m1 m2
2
m1
m1d
m 2d
m1 m2 m1 m2
C
d
Gm1m2
d2
G(m1 m2 )
d3
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or Period of revolution
T
d3
2
G( m1 m 2 )
Ratio of Angular Momenta is
J1
J2
I1
I2
I1
I2
m1
m2
m2 d
m1 m2
m1d
m1 m2
2
2
m2
m1
Ratio of kinetic energies is
K1
K2
7.
1
I
2 1
1
I
2 2
2
2
I1
I2
m2
m1
MOTION OF A SATELLITE IN ELLIPTICAL PATH
Whenever a satellite is in a circular or elliptical path, these orbits are called bounded orbits as satellite
is moving in an orbit bounded to earth. The bound nature of orbit means that the kinetic energy of
satellite is not enough at any point in the orbit to take the satellite to infinity. In equation shown
negative total energy of a revolving satellite shows its boundness to earth. Even when a body is in
elliptical path around the earth, its total energy must be negative. Lets first discuss how a satellite or
a body can be in elliptical path.
Consider a body (satellite) of mass m in a circular path of radius r around the earth as shown in
figure. we've discussed that in circular path the net gravitational frame on body is exactly balancing
the centrifugal force on it in radial direction relative to a rotating frame with the body.
Fe
m
v
FG
r
Me
Re
C
path-I
path-II
If suddenly the velocity of body decreases then the centrifugal force on it becomes less then the
gravitational force acting on it and due to this it can not continue in the circular orbit and will come
inward from the circular orbit due to unbalanced force. Mathematical analysis shown that this path-I
along which the body is now moving is an ellipse. The analytical calculations of the laws for this path is
beyond the scope of this book. But it should be kept in mind that if velocity of a body at a distance r
from earth's centre tangential to the circular orbit is less than
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then its path will be elliptical
r
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with earth centre at one of the foci of the ellipse.
GMe
then it must move out of the circular path due to
r
unbalancing of forces again but this time Fe > Fg. Due to this if speed of body is not increased by such
a value that its kinetic energy can take the particle to infinity then it will follow in a bigger elliptical
orbit as shown in figure in path-II, with earth's at one of the foci of the orbit.
Similarly if the speed of body exceeds
GMe
and the
r
speed is decreased to such a value that the elliptical orbit will intersect the earth's surface as shown
in figure, then body will follow an arc of ellipse and will fall back to earth.
In above case when speed of body was decreased and its value is lesser than
v0
v<v0
r
GM e
r
arc of
ellipse
Me
Re
C
7.2
Satellite Motion and Angular Momentum Conservation
We've discussed that when a body is in bounded orbit around a planet it can be in circular or elliptical
path depending on its kinetic energy at the time of launching. Lets consider a case when a satellite is
launched in an orbit around the earth.
A satellite S is first fired away from earth source in vertical direction to penetrate the earth's
atmosphere. When it reaches point A, it is imparted a velocity in tangential direction to start its
revolution around the earth in its orbit.
v
r
v1>v0
Fg
Me
A
S
C
r1
B
r2
Earth
v2
This velocity is termed as insertion velocity, if the velocity imparted to satellite is v0
GMe
then it
r1
starts following the circular path shown in figure. If velocity imparted is v1 > v0 then it will trace the
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elliptical path shown. During this motion the only force acting on satellite is the gravitational force due
to earth which is acting along the line joining satellite and centre of earth.
As the force on satellite always passes through centre of earth during motion, we can say that on
satellite there is no torque acting about centre of earth thus total angular momentum of satellite
during orbital motion remains constant about earth's centre.
As no external force is involved for earth-satellite system, no external work is being done here so we
can also state that total mechanical energy of system also remains conserved.
In the elliptical path of satellite shown in figure if r1 and r2 are the shortest distance (perigee) and
farthest distance (appogee) of satellite from earth and at the points, velocities of satellite are v1 and
v2 then we have according to conservation of angular momentum, the angular momentum of satellite
at a general point is given as
L = mv1r1 = mv2r2 = mvr sin
During motion the total mechanical energy of satellite (kinetic + potential) also remains conserved.
Thus the total energy of satellite can be given as
E
GMem
1
mv12 –
2
r1
GMem
1
mv22 –
2
r2
GMem
1
mv 2 –
2
r
Using the above relations in equation written above we can find velocities v1 and v2 of satellite at
nearest and farthest locations in terms of r1 and r2.
7.3
Projection of Satellites and Spaceships From Earth
To project a body into space, first it should be taken to a
height where no atmopshere is present then it is projected
with some initial speed. The path followed by the body also
depends on the projection speed. Lets discuss the cases
step by step.
Consider the situation shown in figure. A body of mass m is
taken to a height h above the surface of earth to a point A
and then projected with an insertion velocity vp as shown
in figure.
If we wish to launch the body as an earth's satellite in
circular path the velocity of projection must be
vp
GMe
Re h
A
vp
h
R
R
Earth
...(1)
If h is small compared to radius of earth, we have
v1 = v p =
GMe
Re
gsR e = 7.93 km/s.
This velocity v1 = 7.93 km/s with which, when a body is thrown from earth's surface tangentially so
that after projection it becomes a satellite of earth in a circular orbit around it, is called orbital speed
or first cosmic velocity.
We've already discussed that if projection speed is lesser the orbital speed, body will start following
the inner ellipse and if velocity of projection is increased the body will follow the outer ellipse. If
projection speed of the satellite is further increased, the outer ellipse will also become bigger and at a
particular higher projection speed, it may also be possible that body will go to infinity and will never
come back to earth again.
We have discussed that negative total energy of body shows its boundness. If we write the total
energy of a body projected from point A as shown in figure is
GMem
1
E
mv2p –
2
Re h
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If after projection body becomes a satellite of earth then it implies it is bounded to earth and its total
energy is negative. If at point A, that much of kinetic energy is imparted to the body so that total
energy of body becomes zero then it implies that the body will reach to infinity and escape from
gravitational field of earth. If vII is such a velocity then we have
GMem
1
mv 2II –
0
2
Re h
or
v II
For h << Re, we have v II
2GMe
Re h
2v1
2GMe
Re
2gsR e
.......(2)
11.2km / s .......(3)
Thus from earth's surface a body is thrown at a speed of 11.2 km/s, it will escape from earth's
gravitation. If the projection speed of body is less than this value total energy of body is negative and
it wil orbit the earth in elliptical orbit. This velocity is referred as the second cosmic velocity or escape
velocity. When a body is thrown with this speed, it follows a parabolic trajectory and will become free
from earth's gravitational attraction.
When body is thrown with speed more then vII then it moves along a hyperbolic trajectory and also
leaves the region where the earth's gravitational attraction acts. Also when it reaches infinity some
kinetic energy will be left in it and it becomes a satellite of sun, that is small artificial planet.
v
v>vII
hyperbolic
trajectory
vII
hyperbolic
trajectory
v<v1
C
inner ellipse
circle
v=v1
vI < v < vII
outer ellipse
All the calculations we've performed till now do not take into account the influence of the sun and of
the planets on the motion of the projected body. In other words we have assumed that the reference
frame connected with the earth is an inertial frame and the body moves relative to it. But in reality the
whole system body and the earth is in a non inertial frame which is permanently accelerated relative to
sun.
Lets take some examples to understand some basic concepts related to gravitational energy and
projection.
Ex.14 A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity
has now to be imparted to the spaceship in the orbit of overcome the gravitational pull. (Radius
of the earth = 6400 km and g = 9.8 m/sec.)
Sol.
In an orbit close to earth's surface velocity of space ship is v
GM
R
gR
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We know escape velocity is vII
2gR
Hence additional velocity required to be imparted is
v = vII – v
( 2 – 1) gR
= ( 2 – 1) 9.8 6400 103 = 3.28 × 103 m/s
Ex.15 A particle is fired vertically upward with a speed of 9.8 km/s. Find the maximum height attained
by the particle. Radius of the earth = 6400 km and g at the surface = 9.8 m/s2. Consider only
earth's gravitation.
Sol.
Initial energy of particle on earth's surface is
1
GMm
mu2 –
2
R
If the particle reaches upto a height h above the surface of earth then its final energy will only be the
gravitational potential energy.
Er
GMm
R h
According to energy conservation, we have
Et = Ef
Ef
or
1
GMm
mu2 –
2
R
or
h
2gR 2
–R
2gR – u2
–
GMm
R h
–
or
1 2
u – gR
2
–
gR 2
R h
2 9.8 (6400 103 )2
– 6400 103 = (27300 – 6400) × 103 = 20900 km
2 9.8 6400 103 – (9.8)2
Ex.16 A satellite of mass m is orbiting the earth in a circular orbit of radius r. It starts losing energy
slowly at a constant rate C due to friction. If Me and Re denote the mass and radius of the earth
respectively, show the the satellite falls on the earth in a limit t given by
G mMe 1 1
–
2C
Re r
Let velocity of satellite in its orbit of radius r be v then we have
t
Sol.
GMe
r
When satellite approaches earth's surface, if its velocity becomes v', then it is given as
v
GMe
Re
The total initial energy of satellite at a distance r is
GMem
1
1 GMem
mv2 –
ETf K f Ur
–
2
Re
2 r
The total final energy of satellite at a distance Re is
GMem
1
1 GMem
mv '2–
–
ETf K f Ur
2
Re
2 Re
As satellite is loosing energy at rate C, if it takes a time t in reaching earth, we have
v'
1
1 1
–
Ct ETi – ETf = GMem
2
Re r
t=
GMem 1 1
–
2C
Re r
Ex.17 An artifical satellite is moving in a circular orbit around the earth with a speed equal to half the
magnitude of escape velocity from the earth.
(i) Determine the height of the satellite above earth's surface.
(ii) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find
the speed with which it hits the surface of the earth.
Sol.
(i) Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then
escape velocity from earth vc =
2gR e
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gR e
2
Further we know orbital speed of satallite at a height h is
velocity of satellite
vs
=
Re2 g
Re h
GMe
r
or
v2s
R 2g
R h
From equation written above, we get
h = R = 6400 km
(ii) Now total energy at height h = total energy at earth's surface (principle of conservation of energy)
or
0 – GMe
or
1
mv2
2
Solving we get
or
8.
m
R
h
1
m
mv2 – GMe
2
Re
GMem GMem
–
Re
2R e
v
[As h = R]
gR e
9.8 6400 103 = 7.919 km/s
COMMUNICATION SATELLITES
Communication satellite around the earth are used by Information Technology for spreading information
through out the globe.
Figure shows as to how using satellites an information from an earth station, located at a point on
earth's surface ca be sent throughout the world.
First the information is sent to the nearest satellite in the range of earth station by means of
electromanetic waves then that satellite broadcasts the signal to the region of earth exposed to this
satellite and also send the same signal to other satellite for broadcasting in other parts of the globe.
8.1
Geostationary Satellite and Parking Orbit
There are so many types of communication satellites revolving around the arth in different orbits at
different heights depending on their utility. Some of which are Geostationary satellites, which appears
at rest relative to earth or which have same angular velocity as that of earth's rotation i.e., with a time
peiod of 24 hr. such satellite must be orbiting in an orbit of specific radius. This orbit is called parking
orbit. If a Geostationary satellite is at a height h above the earth's surface then its orbiting speed is
given as
v gs
GMe
(R e h)
The time period of its revolution can be given Kepler's third law as
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or
or
T2
4 2
(R e
GMe
h)3
T2
4 2
(R e
gsR e2
h)3
h
gsR 2e 2
T
4 2
1/ 3
– Re
1/ 3
9.8 [6.4 106 ] [86400]2
– 6.4 106 = 35954.6 km ~– 36000 km
or
4 (3.14)2
Thus when a satellite is launched in an orbit at a height of about 36000 km above the quator then it will
appear to be at rest with respect to a point on Earth's surface. A Geostationary satellite must have in
orbit in equatorial plane due to the geographic limitation because of irregular geometry of earth
(ellipsoidal shape.)
In short
Plane of the satellite should pass through centre of the planet
For geostationary satallites plane should be equatorial plane
Time peirod should be 24 hrs & direction should be west to east
For any point on the earth, geostationary satellite is stationary.
h
•
•
•
•
8.2
Broadcasting Region of a Satellite
Now as we known the height of a geostationary satellite we can easily find the area of earth exposed
to the satellite or area of the region in which the comunication can be mode using this satellite.
Figure shown earth and its exposed area to a geostationary satellite. Here the angle can be given as
Re
Re h
cos–1
Axis of rotation
of earth
Now we can find the solid angle
subtend on earth's centre as
= 2 (1 – cos )
2
1–
Re
Re h
which the exposed area
h
Re
2 h
Re h
Thus the area of earth's surface to geostationary satellite is
S
Re
Earth
2 hR 2e
Re h
R 2e
Lets take some examples to understand the concept in detail.
Ex.18 A satellite is revolving around the earth in an orbit of radius double that of the parking orbit and
revolving in same sense. Find the periodic time duration between two instants when this
satallite is closest to a geostationary satellite.
Sol.
We know that the time period of revolution of a satellite is given as
4 2 3
r
[Kepler's III law]
GMe
For satellite given in problem and for a geostationary satellite we have
T2
3
3
T1
r1
r1
T1
T2 = (2)3 × 24 = 192 hr
or
T2
r2
r2
If t be the time between two sucessive instants when the satellite are closed then we must have
2
– 1
1
2
2
and 2 are the angular speeds of the two planets
t
Where
1
2
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Ex.19 Find the minimum colatitude which can directly receive a signal from a geostationary satellite.
Sol.
The farthest point on earth, which can receive
signals from the parking orbit is the point
P
where a length is drawn on earth surface from
satellite as shown in figure.
h
R
The colatitude of point P can be obtained
from figure as
Parking orbit
Re
1
~
sin = R
–
h
7
e
We known for a parking orbit h ~– 6Re
AOR
1
Thus we have
7
Ex.20 If a satellite is revolving around the earth in a circular orbit in a plane containing earth's axis of
rotation. if the angular speed of satellite is equal to that of earth, find the time it takes to move
from a point above north pole of a point above the equator.
orbit
Sol.
A satellite which rotates with angular speed equal to earth's rotation
N
has an orbit radius 7 Re and the angular speed of revolution is
sin–1
2
T
2
86400
7.27 10 –5rad / s
Re
When satellite moves from a point above north pole to a point
above equator, it traverses an angle /2, this time taken is
/2
t
S
= 21600 s = 6 hrs.
AOR
Ex.21 A satellite is orbiting around the earth in an orbit in equatorial plane of radius 2Re where Re is
the radius of earth. Find the area on earth, this satellite covers for communication purpose in
its complete revolution.
Sol.
As shown in figure when statelite S revolves, it
patch-1
N
covers a complete circular belt on earth's surface
P
for communication. If the colatitude of the farthest
point on surface upto which singals can be received
2R e
(point P) is then we have
C Re
sin
Re
2R e
1
2
or
=
6
During revolution satellite leaves two spherical patches 1 and 2 on earth surface
at north and south
S
patch-2
poles where no signals can be transmitted due to curvature. The areas of these patches can be
obtained by solid angles.
AOR
The solid angle subtended by a patch on earth's centre is
= 2 (1 – cos ) =
(2 –
3 ) st.
Area of patch 1 and 2 is
AP
R 2e
( 2 – 3 )R2e
Thus total area on earth's surface to which communication can be made is
AC
4 R 2e – 2AP
4 R2e – 2 ( 2 – 3 )R2e
2 R2e (2 – 2
3 ) = 2 3 R 2e
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9.
Page # 25
KEPLER'S LAWS OF PLANETARY MOTION
The motions of planet in universe have always been a puzzle. In 17th century Johannes Kepler, after a
life time of study worded out some empirical laws based on the analysis of astronomical measurements
of Tycho Brahe. Kepler formulated his laws, which are kinematical description of planetary motion. Now
we discuss these laws step by step.
9.1
Kepler's First Law [The Law of Orbits]
Kepler's first law is illustrated in the image shown in figure. It states that "All the planets move around
the sun in ellipitcal orbits with sun at one of the focus not at centre of orbit."
It is observed that the orbits of planets around sun are very less ecentric or approximately circular
Planet
Sun
9.2
Focus
Kepler's Second Law [The Law of Areas]
Kepler's second Law is basically an alternative statement of law of conservation of momentum. It is
illustrated in the image shown in figure(a). We know from angular momentum conservation, in elliptical
orbit plane will move faster when it is nearer to the sun. Thus when a planet executes elliptical orbit its
angular speed changes continuously as it moves in the orbit. The point of nearest approach of the
planet to the sun is termed perihelion. The point of greatest seperation is termed aphelion. Hence by
angular momentum conservation we can state that the planet moves with maximum speed when it is
near perihelion and moves with slowest speed when it is near aphelion.
v
A
Perihelion v
1
r1
Sun
D
d
r
C
E
v2
Aphelion
r2
Sun
B
S
(b)
(a)
Kepler's second law states that "The line joining the sun and planet sweeps out equal areas in equal
time or the rate of sweeping area by the position vector of the planet with respect to sun remains
constant. "This is shown in figure (b).
The above statement of Kepler's second law can be verified by the law of conservation of angular
momentum. To verify this consider the moving planet around the sun at a general point C in the orbit at
speed v. Let at this instant the distance of planet from sun is r. If be the angle between position
vector r of planet and its velocity vector then the angular momentum of planet at this instant is
L = m v r sin
...(1)
In an elemental time the planet will cover a small distance CD = dl and will travel to another adjacent
point D as shown in figure (a), thus the distance CD = vdt. In this duration dt, the position vector r
sweeps out an area equal to that of triangle SCD, which is calculated as
Area of triangle SCD is
dA =
=
1
× r × vdt sin ( – )
2
1
r v sin . dt
2
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Thus the rate of sweeping area by the position vector r is
dA
dt
1
rv sin
2
dA
dt
L
2m
Now from equation (1)
cons tan t
...(2)
The expression in equation (2) verifies the statement of Kepler' II law of planetary motion.
9.3
Kepler's Third law [The Law of Periods]
Kepler's Third Law is concerned with the time period of revolution of planets. It states that "The time
period of revolution of a planet in its orbit around the sun is directly proportional to the cube of semimajor axis of the elliptical path around the sun"
If 'T' is the period of revolution and 'a' be the semi-major axis of the path of planet then according to
Kepler's III law, we have
T2
a3
For circular orbits, it is a special case of ellipse when its major and minor axis are equal. If a planet is in
a circular orbit of radius r around the sun then its revolution speed must be given as
v
GMs
r
Where Ms is the mass of sun. Here you can recall that this speed is independent from the mass of
planet. Here the time period of revolution can be given as
T
T
2 r
v
or
2 r
GMs
r
Squaring equation written above, we get
4 2 3
r
...(1)
GMs
Equation (1) verifies the statement of Kepler's third law for circular orbits. Similarly we can also verify
it for elliptical orbits. For this we start from the relation we've derived earlier for rate of sweeping area
by the position vector of planet with respect to sun which is given as
T2
dA
dt
L
2m
Ex.22 The moon revolves around the earth 13 times per year. If the ratio of the distance of the earth
from the sun to the distance of the moon from the earth is 392, find the ratio of mass of the sun
to the mass of the earth.
Sol.
The time period Te of earth around sun of mass Ms is given by
Te2
4 2
GMs
re3
...(1)
Where re is the radius of the earth.
Similarly, time period Tm of moon around earth is given by
Tm2
4 2
GMe
rm3
...(2)
Dividing equation(1) by equation (2), we get
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GRAVITATION
Page # 27
Te
Tm
or
2
Me
Ms
Ms
Me
2
Tm
Te
3
re
rm
re
rm
3
...(3)
Substituting the given values, we get
Ms
Me
(13)
1
2
(392)3
3.56 105
Ex.23 A satellite revolves around a planet in an elliptical orbit. Its maximum and minimum distances
from the planet are 1.5 × 107 m and 0.5 × 107 m respectively. If the speed of the satellite at the
farthest point be 5 × 103 m/s, calculate the speed at the nearest point.
v1
r1
r2
Appogee
Perigee
m
Sol.
v2
In case of elliptical orbit, the speed of satellite varies constantly as shown in figure. Thus according to
the law of conservation of angular momentum, the satellite must move faster at a point of closest
approach (Perigee) than at a farthest point (Appogee).
We know that
L
Hence, at the two points,
r mv
L = m v1 r 1 = m v2 r 2
v1
v2
or
r2
r1
Substituting the given values, we get
5 103
v2
0.5 107
1.5 107
v2 = 1.5 × 104 m/s
Ex.24 Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a
period of revolution T. On what power of r, will the square of time period depend if the
gravitational force of attraction between the planet and the star is proportional to r–5/2.
Sol.
As gravitation provides centripetal force
mv2
r
So that
T=
K
5/2
r
2 r
v
,
2 r
i.e.,
mr 3 / 2
K
v2
or
K
mr3 / 2
T2
4 2m 7 /2
r ;
K
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Ex.25 A satellite is revolving round the earth in a circular orbit of
radius a with velocity v0. A particle is projected from the
satellite in forward direction with relative velocity
v+v0
Me
v ( 5 / 4 – 1) v0 . Calculate, during subsequent motion of
the particle its minimum and maximum distances from earth's
centre.
The corresponding situation is shown in figure.
v0
Initial velocity of satellite
a
C
GM
a
When particle is thrown with the velocity v relative to
satellite, the resultant velocity of particle will become
vR = v0 + v
=
5
v0
4
Re
r
v1
5 GM
4 a
As the particle velocity is greater than the velocity required for circular orbit, hence the particle path
deviates from circular path to elliptical path. At position of minimum and maximum distance velocity
vectors are perpendicular to instantaneous radius vector. In this elliptical path the minimum distance of
particle from earth's centre is a and maximum speed in the path is vR and let the maximum distance and
minimum speed in the path is r and v1 respectively.
Now as angular momentum and total energy remain conserved. Applying the law of conservation
of angular momentum, we have
m v1 r = m(v0 + v) a
[m = mass of particle]
v1
or
( v0
v)a
r
a
r
5 GM
4 a
1
= r
5
GMa
4
Applying the law of conservation of energy
1
GMm
2
mv1 –
2
r
or
or
1
m(v 0
2
2
v) –
GMm
a
1
5 GMa
G Mm
m
–
2
4 r2
r
1
5 GM
GM m
m
–
2
4 a
a
5
8
–
a 1
–
r2 r
5
8
1 1
–
a a
3
8a
3r2 – 8 ar + 5 a2 = 0
or
r = a or
5a
3
Thus minimum distance of the particle = a
And maximum distance of the particle =
5a
3
Ex.26 A sky lab of mass 2 × 103 kg is first launched from the surface of earth in a circular orbit of radius
2 R (from the centre of earth) and then it is shifted from this circular orbit to another circular
orbit of radius 3 R. Calculate the minimum energy required (a) to place the lab in the first orbit
(b) to shift the lab from first orbit to the second orbit. Given, R = 6400 km and g = 10 m/s2.
Sol.
(a) The energy of the sky lab on the surface of earth
ES = KE + PE = 0 + –
GMm
GMm
=–
R
R
And the total energy of the sky lab in an orbit of radius 2 R is
E1
–
GMm
4R
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So the energy required to placed the lab from the surface of earth to the orbit of radius 2R is given as
E1 – Es = –
or
E
3m
4 R
GMm
GMm
– –
4R
R
gR 2
3 GMm
4 R
3
mgR
4
As g
GM
R2
3
3
(2 × 103 × 10 × 6.4 × 106) = (12.8 × 1010) = 9.6 × 1010 J
4
4
(b) As for II orbit of radius 3R the total energy of sky lab is
or
E=
E2
or
or
E2 – E1 = –
E=
–
GMm
2(3R)
–
GMm
6R
GMm
GMm
– –
6R
4R
1 GMm
12 R
1
1
mgR =
(12.8 1010 ) = 1.1 × 1010 J
12
12
Ex.27 A satellite is revolving around a planet of mass M in an elliptic orbit of semimajor axis a. Show
that the orbital speed of the satellite when it is at a distance r from the focus will be given by :
v2
Sol.
GM
2 1
–
r a
As in case of elliptic orbit with semi major axes a, of a satellite total mechanical energy remains
constant, at any position of satellite in the orbit, given as
E
–
GMm
2a
KE + PE = –
or
GMm
2a
...(1)
Now, if at position r, v is the orbital speed of satellite, we have
1
GMm
mv2 and PE = –
2
r
So from equation (1) and (2), we have
KE =
GMm
1
GMm
mv2 –
=–
, i.e.,,
2a
2
r
...(2)
v2
GM
2 1
–
r a
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OBJECTIVE PROBLEMS (JEE MAIN)
Exercise - I
1. If R is the radius of the earth and g the
acceleration due to gravity on the earth’s surface,
the mean density of the earth is
(A) 4 G/3gR
(B) 3 R/4gG
(C) 3g/4 RG
(D) Rg/12G
Sol.
Sol.
4. At what altitude will the acceleration due to gravity
be 25% of that at the earth’s surface (given radius of
earth is R) ?
(A) R/4
(B) R
(C) 3R/8
(D) R/2
Sol.
2. The height above surface of earth where the value
of gravitational acceleration is one fourth of that at
surface, will be
(A) Re/4
(B) Re/2
(C) 3Re/4
(D) Re
Sol.
5. If the rotational motion of earth increases, then
the weight of the body (A) will remain same
(B) will increase
(C) will decrease
(D) none of these
Sol.
3. The decrease in the value of g on going to a height
R/2 above the earth’s surface will be (A) g/2
(B)
5g
9
(C)
4g
9
(D)
g
3
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6. On doubling the distance between two masses the
gravitational force between them will (A) remain unchanged
(B) become one-fourth
(C) become half
(D) become double
Sol.
Sol.
9. If the radius of the earth be increased by a factor
of 5, by what factor its density be changed to keep
the value of g the same ?
(A) 1/25 (B) 1/5
(C) 1 / 5 (D) 5
Sol.
7. If the acceleration due to gravity inside the earth
is to be kept constant, then the relation between the
density d and the distance r from the centre of earth
will be (A) d
r
(B) d
r1/2
(C) d
1/r
(D) d
1
r2
Sol.
10. The mass and diameter of a planet are twice
those of earth. What will be the period of oscillation
of a pendulum on this planet if it is a seconds pendulum
on earth ?
(A)
(C)
8. Let be the angular velocity of the earth’s rotation
about its axis. Assume that the acceleration due to
gravity on the earth’s surface has the same value at
the equator and the poles. An object weighed at the
equator gives the same reading as a reading taken at
a depth d below earth’s surface at a pole (d << R).
The value of d is
2
(A)
R2
(B)
g
2 2
R
2 2R2
(C)
2g
g
(D)
2 second
1
2
second
(B) 2 2 seconds
(D)
1
2 2
second
Sol.
Rg
g
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11. A particle of mass M is at a distance a from
surface of a thin spherical shell of equal mass and
having radius a.
M
M
13. The escape velocity from a planet is v0. The
escape velocity from a planet having twice the radius
but same density will be (A) 0.5 v0
(B) v0
(C) 2v0
(D) 4v0
Sol.
a
(A) Gravitational field and potential both are zero at
centre of the shell
(B) Gravitational field is zero not only inside the shell
but at a point outside the shell also
(C) Inside the shell, gravitational field alone is zero
(D) Neither gravitational field nor gravitational potential
is zero inside the shell
Sol.
14. Two planets A and B have the same material
density. If the radius of A is twice that of B, then the
vA
ratio of the escape velocity v is
B
(A) 2
(B) 2
(C) 1 / 2 (D) 1/2
Sol.
12. If the kinetic energy of a satellite orbiting around
the earth is doubled then (A) the satellite will escape into the space.
(B) the satellite will fall down on the earth
(C) radius of its orbit will be doubled
(D) radius of its orbit will become half.
Sol.
15. A hollow spherical shell is compressed to half its
radius. The gravitational potential at the centre
(A) increases
(B) decreases
(C) remains same
(D) during the compression increases then returns at
the previous value
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Sol.
Sol.
18. A satellite revolves in the geostationary orbit but
16. A (nonrotating) star collapses onto itself from an
initial radius Ri with its mass remaining unchanged.
Which curve in figure best gives the gravitational
acceleration ag on the surface of the star as a function
of the radius of the star during the collapse ?
in a direction east to west. The time interval between
its successive passing about a point on the equator is
(A) 48 hrs
(B) 24 hrs
(C) 12 hrs
(D) never
Sol.
ag
b d
a
c
Ri
(A) a
Sol.
(B) b
(C) c
R
(D) d
19. Two point masses of mass 4m and m respectively
separated by d distance are revolving under mutual
force of attraction. Ratio of their kinetic energies will
be
(A) 1 : 4 (B) 1 : 5(C) 1 : 1 (D) 1 : 2
Sol.
17. A satellite of the earth is revolving in circular
orbit with a uniform velocity V. If the gravitational
force suddenly disappears, the statellite will
(A) continue to move with the same velocity in the
same orbit
(B) move tangentially to the original orbit with velocity V
(C) fall down with increasing velocity
(D) come to a stop somewhere in its original orbit
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20. Select the correct choice(s) :
(A) The gravitational field inside a spherical cavity,
Sol.
within a spherical planet must be non zero and uniform.
(B) When a body is projected horizontally at an
appreciable large height above the earth, with a
velocity less than for a circular orbit, it will fall to the
earth along a parabolic path
(C) A body of zero total mechanical energy placed in
a gravitational field will escape the field
(D) Earth’s satellite must be in equatorial plane.
Sol.
22. Two different masses are dropped from same
heights, then just before these strike the ground, the
following is same :
(A) kinetic energy
(B) potential energy
(C) linear momentum
(D) Acceleration
Sol.
21. The figure shows the variation of energy with the
orbit radius of a body in circular planetary motion.
Find the correct state ment about the
curves A, B and C
A
C
23. When a satellite moves around the earth in a
certain orbit, the quantity which remains constant is
(A) angular velocity
(B) kinetic energy
(C) aerial velocity
(D) potential energy
Sol.
B
(A) A shows the kinetic energy, B the total energy
and C the potential energy of the system
(B) C shows the total energy, B the kinetic energy
and A the potential energy of the system
(C) C and A are kinetic and potential energies
respectively and B is the total energy of the system
(D) A and B are kinetic and potential energies and C is
the total energy of the system
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24. A body of mass m rises to height h = R/5 from the
earth’s surface, where R is earth’s radius. If g is acceleration due to gravity at earth’s surface, the increase in potential energy is
(A) mg/h
(B)
5
mgh
6
(C)
3
mgh
5
(D)
Sol.
6
mgh
7
Sol.
25. A planet has mass 1/10 of that of earth, while
radius is 1/3 that of earth. If a person can throw a
stone on earth surface to a height of 90m, then he
will be able to throw the stone on that planet to a
height
(A) 90m
(B) 40 m
(C) 100 m
(D) 45 m
Sol.
27. The potential energy of a body of mass 3kg on
the surface of a planet is 54 joule. The escape velocity will be (A) 18m/s
(B) 162 m/s
(C) 36 m/s
(D) 6 m/s
Sol.
28. A body of mass m is situated at a distance 4Re
above the earth’s surface, where Re is the radius of
earth. How much minimum energy be given to the
body so that it may escape (A) mgRe
(B) 2mgRe
(C)
mgR e
5
(D)
mgR e
16
Sol.
26. Work done in taking a body of mass m to a height
nR above s urface of eart h will b e :
(R = radius of earth)
(A) mgnR
(B) mgR (n/n + 1)
(C) mgR
(n 1)
n
(D)
mgR
n(n 1)
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29. A satellite of earth is moving in its orbit with a
constant speed v. If the gravity of earth suddenly
vanishes, then this satellite will (A) continue to move in the orbit with velocity v.
(B) start moving with velocity v in a direction tangential to the orbit
(C) fall down with increased velocity
(D) be lost in outer space.
Sol.
30. The ratio of distances of satellites A and B from
the centre of the earth is 1.4 : 1, then the ratio of
energies of satellites B and A will be –
(A) 1.4 : 1
(B) 2 : 1
(C) 1 : 3
(D) 4 : 1
Sol.
32. A body is dropped by a satellite in its geo-stationary orbit (A) it will burn on entering into the atmosphere.
(B) it will remain in the same place with respect to
the earth.
(C) it will reach the earth in 24 hours
(D) it will perform uncertain motion
Sol.
33. The orbital velocity of an artificial satellite in a
circular orbit just above the earth’s surface is . For a
satellite orbiting at an altitude of half of the earth’s
radius, the orbital velocity is –
(A)
3
2
(B)
3
2
(C)
2
3
(D)
2
3
Sol.
31. A satellite can be in a geostationary orbit around
earth at a distance r from the centre. If the angular
velocity of earth about its axis doubles, a satellite
can now be in a geostationary orbit around earth if its
distance from the center is
r
r
r
r
(A)
(B)
(C)
(D)
2 2
(4)1/ 3
(2)1/ 3
2
Sol.
34. Two artificial satellites A and B are at a distances
rA and r B above the earth’s surface. If the radius of
earth is R, then the ratio of their speeds will be rB R
(A)
rA R
(C)
rB
rA
1/ 2
rB R
(B)
rA R
2
(D)
rB
rA
2
1/ 2
Sol.
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35. A planet of mass m is in an elliptical orbit about
the sun (m << Msun) with an orbital period T. If A be
the area of orbit, then its angular momentum would
be
(A)
2mA
T
(B) mAT
(C)
mA
2T
38. The period of revolutions of two satellites are 3
hours and 24 hours. The ratio of their velocities will
be (A) 1 : 8
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1
Sol.
(D) 2mAT
Sol.
36.If the distance between sun and earth is made 3
times of the present value then gravitational force
between them will become :
(A) 9 times
(C)
1
times
3
(B)
1
times
9
(D) 3 times
Sol.
39. Statement - I : Assuming zero potential at infinity,
gravitational potential at a point cannot be positive.
Statement - 2 : Magnitude of gravitational force
between two particle has inverse square dependence
on distance between two particles.
(A) Statement - 1 is true, statement-2 is true and statement-2 is correct explanation for statement-1
(B) Statement -1 is true, statement-2 is true and
statement - 2 is NOT the correct explanation for statement-1
(C) Statement - 1 is true, statement - 2 is false.
(D) Statement - 1 is false, statement - 2 is true.
Sol.
37. If a body is carried from surface of earth to moon,
then
(A) the weight of a body will continuously increase
(B) the mass of a body will continuously increase
(C) the weight of a body will decrease first, become
zero and then increase,
(D) the mass of a body will decrease first, become
zero and then increase.
Sol.
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Exercise - II
(SINGLE CORRECT)
1. Two masses m1 & m2 are initially at rest and are
separated by a very large distance. If the masses
approach each other subsequently, due to gravitational
attraction between them, their relative velocity of
approach at a separation distance of d is
3. A spherical uniform planet is rotating about its axis.
The velocity of a point on its equator is V. Due to the
rotation of planet about its axis the acceleration due
to gravity g at equator is 1/2 of g at poles. The escape
velocity of a particle on the pole of planet in terms of
V.
2Gd
(A) ( m m )
1
2
(A) Ve = 2V (B) Ve = V (C) Ve = V/2 (D) Ve =
(C) (m1 m2 )
(B)
2G
d
( m1 m2 ) G
2d
Sol.
1/ 2
(D) (m1 + m2)
1/2
3V
2Gd
Sol.
2. A man of mass m starts falling towards a planet of
mass M and radius R. As he reaches near to the surface,
he realizes that he will pass through a small hole in
the planet. As he enters the hole, he sees that the
planet is really made of two pieces a spherical shell of
negligible thickness of mass 2M/3 and a point mass M/
3 at the centre. Change in the force of gravity
experienced by the man is
(A)
2 GMm
3 R2
(B) 0
(C)
1 GMm
3 R2
(D)
4. The escape velocity for a planet is ve. A tunnel is
dug along a diameter of the planet and a small body is
dropped into it at the surface. When the body reaches
the centre of the planet, its speed will be
(A) ve
(B)
ve
2
(C)
ve
2
(D) zero
Sol.
4 GMm
3 R2
Sol.
5. If a tunnel is cut at any orientation through earth,
then a ball released from one end will reach the other
end in time (neglect earth rotation)
(A) 84.6 minutes
(B) 42.3 minutes
(C) 8 minutes
(D) depends on orientation
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Sol.
Paragraph Q. 6 & Q. 7
Two uniform spherical stars made of same material
have radii R and 2R. Mass of the smaller planet is m.
They start moving from rest towards each other from
a large distance under mutual force of gravity. The
collision between the stars is inelastic with coefficient
of restitution 1/2.
6. Kinetic energy of the system just after the collision
is
8Gm2
2Gm2
(B)
3R
3R
(D) cannot be determined
Sol.
(A)
(C)
7. The maximum separation between their centres
after their first collision
(A) 4R
(B) 6R
(C) 8R
(D) 12R
Sol.
4Gm2
3R
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Page # 40
8. A mass is at the center of a square, with four
masses at the corners as shown.
5M
3M
M
(A)
2M
3M
M
(B)
M
5M
M
2M
5M
3M
2M
3M
2M
(C)
M
2M
(D)
5M
Sol.
2M
M
Rank the choices according to the magnitude of the
gravitational force on the center mass.
(A) FA = FB < FC = FD (B) FA > FB < FD < FC
(C) FA = FB > FC = FD (D) none
Sol.
10. A satellite of mass 5M orbits the earth in a circular
orbit. At one point in its orbit, the satellite explodes
into two pieces, one of mass M and the other of mass
4M. After the explosion the mass M ends up travelling
in the same circular orbit, but in opposite direction.
After explosion the mass 4M is in
(A) bound orbit
(B) unbound orbit
(C) partially bound orbit
(D) data is insufficient to determine the nature of the orbit
Sol.
9. A satellite of mass m, initially at rest on the earth,
is launched into a circular orbit at a height equal to
the radius of the earth. The minimum energy required
is
(A)
3
mgR
4
(B)
1
mgR
2
(C)
1
3
mgR (D) mgR
4
4
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GRAVITATION
Page # 41
Q. 11 & 12
Figure shows the orbit of a planet P round the sun S.
AB and CD are the minor and major axes of the ellipse.
A
Sol.
P
D
C
S
B
11. If t1 is the time taken by the planet to travel
along ACB and t2 the time along BDA, then
(A) t1 = t2
(B) t1 > t2
(C) t1 < t2
(D) nothing can be concluded
14. Figure shows the variation of energy with the
orbit radius r of a satellite in a circular motion. Select
the correct statement.
Sol.
X
Z
Y
(A) Z is total energy, Y is kinetic energy and X is
potential energy
(B) X is kinetic energy, Y is total energy and Z is
potential energy
12. If U is the potential energy and K kinetic energy
then |U| > |K| at
(A) Only D
(B) Only C
(D) neither D nor C
(C) both D & C
(C) X is kinetic energy, Y is potential energy and Z is
total energy
(D) Z is kinetic energy, X is potential energy and Y is
total energy
Sol.
Sol.
13. Satellites A and B are orbiting around the earth in
orbits of ratio R and 4R respectively. The ratio of their
areal velocities is
(A) 1 : 2
(B) 1 : 4
(C) 1 : 8
(D) 1 : 16
(MULTIPLE CORRECT)
15. Assuming the earth to be a sphere of uniform
density the acceleration due to gravity
(A) at a point outside the earth is inversely proportional to the square of its distance from the center
(B) at a point outside the earth is inversely proportional to its distance from the centre
(C) at a point inside is zero
(D) at a point inside is proportional to its distance
from the centre
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Page # 42
Sol.
Sol.
16. Two masses m1 and m2 (m1 < m2) are released
from rest from a finite distance. They start under their
mutual gravitational attraction
(A) acceleration of m1 is more than that of m2
(B) acceleration of m2 is more than that of m1
(C) centre of mass of system will remain at rest in all
the references frame
(D) total energy of system remains constant
Sol.
18. A geostationary satellite is at a height h above
the surface of earth. If earth radius is R
R
R
h
(A) The minimum colatitude on earth upto which the
satellite can be used for communication is sin–1 (R/R + h)
(B) The maximum colatitudes on earth upto which the
satellite can be used for communication is sin–1 (R/R + h)
(C) The area on earth escaped from this satellite is
given as 2 R2(1 + sin )
(D) The area on earth escaped from this satellite is
given as 2 R2(1 + cos )
Sol.
17. In side a hollow spherical shell
(A) everywhere gravitational potential is zero
(B) everywhere gravitational field is zero
(C) everywhere gravitational potential is same
(D) everywhere gravitational field is same
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GRAVITATION
19. When a satellite in a circular orbit around the
earth enters the atmospheric region, it encounters
small air resistance to its motion. Then
(A) its kinetic energy increases
(B) its kinetic energy decreases
(C) its angular momentum about the earth decreases
(D) its period of revolution around the earth increases
Sol.
20. A communications Earth satellite
(A) goes round the earth from east to west
(B) can be in the equatorial plane only
(C) can be vertically above any place on the earth
(D) goes round the earth from west to east
Sol.
21. An earth satellite is moved from one stable circular
orbit to another larger and stable circular orbit. The
following quantities increase for the satellite as a result
of this change
(A) gravitational potential energy
(B) angular velocity
(C) linear orbital velocity
(D) centripetal acceleration
Page # 43
Sol.
22. A satellite S is moving in an elliptical orbit around
the earth. The mass of the satellite is very small
compared to the mass of the earth
(A) the acceleration of S is always directed towards
the centre of the earth
(B) the angular momentum of S about the centre of
the earth changes in direction, but its magnitude
remains constant
(C) the total mechanical energy of S varies periodically
with time
(D) the linear momentum of S remains constant in
magnitude
Sol.
23. If a satellite orbits as close to the earth’s surface
as possible,
(A) its speed is maximum
(B) time period of its rotation is minimum
(C) the total energy of the ‘earth plus satellite’ system
is minimum
(D) the total energy of the ‘earth plus satellite’ system
is maximum
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Page # 44
Sol.
24. For a satellite to orbit around the earth, which of
the following must be true ?
(A) It must be above the equator at some time
(B) It cannot pass over the poles at any time
(C) Its height above the surface cannot exceed 36,000 km
(D) Its period of rotation must be
is radius of earth
Sol.
2
R / g where R
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GRAVITATION
Page # 45
(JEE ADVANCED)
Exercise - III
1. Four masses (each of m) are
placed at the vertices of a regular
pyramid (triangular base) of side ‘a’.
Find the work done by the system
m
while taking them apart so that they
form the pyramid of side ‘2a’.
a
Sol.
m
m
m
2. A small mass and a thin uniform rod each of mass
‘m’ are positioned along the same straight line as shown.
Find the force of gravitational attraction exerted by
the rod on the small mass.
2L
L
m
m
3. An object is projected vertically upward from the
surface of the earth of mass M with a velocity such
that the maximum height reached is eight times the
radius R of the earth. Calculate :
(i) the initial speed of projection
(ii) the speed at half the maximum height.
Sol.
4. A satellite close to the earth is in orbit above the
equator with a period of rotation of 1.5 hours. If it is
above a point P on the equator at some time, it will be
above P again after time ___________.
Sol.
Sol.
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Page # 46
5. A satellite is moving in a circular orbit around the
earth. The total energy of the satellite is E = –2 ×105
J. The amount of energy to be imparted to the satellite
to transfer it to a circular orbit where its potential
energy is U = –2×105J is equal to ___________.
Sol.
7. A point P lies on the axis of a fixed ring of mass M
and radius a, at a distance a from its centre C. A
small particle starts from P and reaches C under
gravitational attraction only. Its speed a C will be
________.
Sol.
8. Calculate the distance from the surface of the earth
at which above and below the surface acceleration
due to gravity is the same.
Sol.
6. Find the gravitational field strength and potential
at the centre of arc of linear mass density subtending
an angle 2 at the centre.
2
Sol.
R
9. Consider two satellites A and B of equal mass m,
moving in the same circular orbit of radius r around
the earth E but in opposite sense of rotation and
therefore on a collision course (see figure).
A
r
B
Me
(a) In terms of G, Me, m and r find the total mechanical
energy EA + EB of the two satellite plus earth system
before collision.
(b) If the collision is completely inelastic so that
wreckage remains as one piece of tangle d material
(mass = 2m), find the total mechanical energy
immediately after collision.
(c) Describe the subsequent motion of the wreckage.
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GRAVITATION
Page # 47
Sol.
12. Find the potential energy of a system of eight
particles placed at the vertices of a cube of side L.
Neglect the self energy of the particles.
Sol.
10. A particle is fired vertically from the surface of
the earth with a velocity k e, where e is the escape
velocity and k <1. Neglecting air resistance and
assuming earth’s radius as Re. Calculate the height to
which it will rise from the surface of the earth.
Sol.
13. A hypothetical planet of
mass M has three moons each
of equal mass ‘m’ each revolving
in the same circular orbit of
radiu s R. The masses are
R
m
equally spaced and thus form
an equilateral triangle. Find
(i) the total P.E. of the system
(ii) the orbital speed of each moon such that they
maintain this configuration.
Sol.
11. A satellite of mass m is orbiting the earth in a
circular orbit of radius r. It starts losing energy due to
small air resistance at the rate of C J/s. Then the time
taken for the satellite to reach the earth is ________.
Sol.
14. Two small dense stars rotate about their common
centre of mass as a binary system with the period
1year for each. One star is of double the mass of the
other and the mass of the lighter one is 1/3 of the
mass of the sun. Find the distance between the stars
if distance between the earth & the sun is R.
Sol.
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15. A sphere of radius R has its centre at the origin.
It has a uniform mass density 0 except that there is a
spherical hole of radius r = R/2 whose centre is at x =
R/2 as in fig. (a) Find gravitational field at points on
the axis for x > R
O
x
17. A thin spherical shell of total mass M and radius R
is held fixed. There is a small hole in the shell. A mass
m is released from rest a distance R from the hole
along a line that passes through the hole and also
through the centre of the shell. This mass subsequently
moves under the gravitational force of the shell. How
long does the mass take to travel from the hole to the
point diametrically opposite.
Sol.
(ii) Show that the gravitational field inside the hole is
uniform, find its magnitude and direction.
Sol.
16. A body moving radially away from a planet of
mass M, when at distance r from planet, explodes in
such a way that two of its many fragments move in
mutually perpendicular circular orbits around the planet.
What will be
(a) then velocity in circular orbits.
(b) maximum distance between the two fragments
before collision and
(c) magnitude of their relative velocity just before
they collide.
Sol.
18. A remote sensing satellite is revolving in an orbit
of radius x the equator of earth. Find the area on
earth surface in which satellite can not send message.
Sol.
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GRAVITATION
Page # 49
1. A satellite P is revolving around the earth at a
height h = radius of earth (R) above equator. Another
satellite Q is at a height 2h revolving in opposite
direction. At an instant the two are at same vertical
line passing through centre of sphere. Find the least
time of after which again they are in this is situation.
Earth
M
PQ
2. A certain triple-star system consists of two stars,
each of mass m, revolving about a central star, mass
M, in the same circular orbit. The two stars stay at
opposite ends of a diameter of the circular orbit, see
figure. Derive an expression for the period of revolution
of the stars; the radius of the orbit is r.
5. A ring of radius R is made from a thin wire of radius
r. If is the density of the material of wire then what
will be the gravitational force exerted by the ring on
the material particle of mass m placed on the axis of
ring at a distance x from its centre. Show that the
force will be maximum when x R / 2 and the maximum
value of force will be given as Fmax
4
2
Gr 2 m
( 3) 3 / 2 R
6. A man can jump over b = 4m wide trench on earth.
If mean density of an imaginary planet is twice that of
the earth, calculate its maximum possible radius so
that he may escape from it by jumping. Given radius
of earth = 6400 km.
7. A launching pad with a spaceship is moving along a
circular orbit of the moon, whose radius R is triple
that of moon Rm. The ship leaves the launching pad
with a relative velocity equal to the launching pad’s
initial orbital velocity v 0 and the launching pad then
m
r
falls to the moon. Determine the angle with the
horizontal at which the launching pad crashes into
the surface if its mass is twice that of the spaceship
m.
M
m
3. Find the gravitational force of interaction between
the mass m and an infinite rod of varying mass density
such that (x) = /x, where x is the distance from
mass m. Given that mass m is placed at a distance d
from the end of the rod on its axis as shown in figure.
x
d
O
m
( x)
x
4. Inside an isolated fixed sphere of radius R and uniform
density r, there is a spherical cavity of radius R/2
such that the surface of the cavity passes through
the centre of the sphere as in figure. A particle of
mass m is released from rest at centre B of the cavity.
Calculate velocity with which particle strikes the centre
A of the sphere.
A
R
B
R/2
8. A satellite of mass m is in an elliptical orbit around
the earth of mass M(M >>m). The speed of the satellite
6GM
5R
where R = its closest distance to the earth. It is
desired to transfer this satellite into a circular orbit
around the earth of radius equal its largest distance
from the earth. Find the increase in its speed to be
imparted at the apogee (farthest point on the elliptical
orbit).
9. A body is launched from the earth’s surface a an
angl e
= 30º to th e hori zontal at a s peed
at its nearest point to the earth (perigee) is
15
. GM
. Neglecting air resistance and earth’s
R
rotation, find (a) the height to which the body will
rise. (ii) The radius of curvature of trajectory at its
top point.
10. Assume that a tunnel is dug across the earth
(radius = R) passing through its centre. Find the time
a particle takes to reach centre of earth if it is
projected into the tunnel from surface of earth with
speed needed for it to escape the gravitational field
of earth.
v0
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Page # 50
Exercise - IV
PREVIOUS YEAR QUESTIONS
JEE MAIN
LEVEL - I
1. If suddenly the gravitational force of attraction
between earth and a satellite revolving around it
becomes zero, then the satellite will [AIEEE 2002]
(A) continue to move in its orbit with same velocity
(B) move tangentially to the original orbit with the
same velocity
(C) become stationary in its orbit
(D) move towards the earth
Sol.
Sol.
4. Energy required to move a body of mass m from
an orbit of radius 2R to 3R is
[AIEEE 2002]
GMm
2
12 R
(A)
(B)
GMm
2
3R
(C)
GMm
GMm
(D)
8R
6R
Sol.
2. The escape velocity of a body depends upon mass
as
[AIEEE 2002]
(A) m0
(B) m1
(C) m2
(D) m3
Sol.
5. The escape velocity for a body projected vertically
upwards from the surface of earth is 11 km/s. If the
body is projected at an angle of 45° with the vertical,
the escape velocity will be
[AIEEE 2003]
(A) 11 2 km / s
(B)
22 km / s
11km / s
(D)
11
m/s
2
(C)
Sol.
3. The kinetic energy needed to project a body of
mass m from the earth's surface (radius R) to infinity
is
[AIEEE 2002]
(A)
mgR
2
(B) 2mgR
(C) mgR
(D)
mgR
4
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GRAVITATION
Page # 51
6. Two spherical bodies of mass M and 5M and radii R
and 2R respectively are released in free space with
initial separation between their centres equal to 12R.
If they attract each other due to gravitational force
only. then the distance covered by the smaller body
just before collision is
[AIEEE 2003]
(A) 2.5 R
(B) 4.5 R
(C) 7.5 R
(D) 1.5 R
Sol.
9. If g is the acceleration due to gravity on the earth's
surface, the gain in the potential energy of an object
of mass m raised from the surface of the earth to a
height equal to the radius R of the earth, is
[AIEEE 2004]
(A) 2 mgR
(B)
1
2
mgR
(C)
1
mgR (D) mgR
4
Sol.
7. The time period of a satellite of earth is 5 h. If the
separation between the earth and the satellite is
increased to 4 times the previous value, the new time
period will become
[AIEEE 2003]
(A) 10 h
(B) 80 h
(C) 40 h
(D) 20 h
Sol.
8. Suppose the gravitational forces varies inversely
as the nth power of distance. Then the time period of
a planet in circular orbit of radius R around the sun will
be proportional to
[AIEEE 2004]
(A)
Sol.
R
n 1
2
(B)
R
n 1
2
(C)
Rn
(D)
R
10.The time period of an earth satellite in circular
orbit is independent of
[AIEEE 2004]
(A) the mass of the satellite
(B) radius of its orbit
(C) both the mass and radius of the orbit
(D) neither the mass of the satellite nor the radius of
its orbit
Sol.
11.A satellite of mass m revolves around the earth of
radius R at a height x from its surface. If g is the
acceleration due to gravity on the surface of the earth,
the orbital speed of the satellite is
[AIEEE 2004]
(A)
gx
n 2
2
gR 2
(C)
R x
(B)
gR
R x
gR2
(D)
R x
1/ 2
Sol.
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Page # 52
12.A particle of mass 10 g is kept on the surface of a
uniform sphere of mass 100 kg and radius 10 cm. Find
the work to be done against the gravitational force
between them, to take the particle far away from the
sphere, (you may take G = 6.67 × 10–11 Nm2/kg–2)
[AIEEE 2005]
(A) 13.34 × 10–10 J
(B) 3.33 × 10–10 J
(C) 6.67 × 10–9J
(D) 6.67 × 10–10 J
Sol.
14.Average density of the earth
(A) does not depend on g
(B) is a complex function of g
(C) is directly proportional to g
(D) is inversely proportional to g
Sol.
13.The change in the value of g at a height h above
the surface of the earth is the same as at a depth d
below the surface of earth. When both d and h are
much smaller than the radius of earth, then which one
of the following is correct ?
[AIEEE 2005]
15. If gE and gM are the accelerations due to gravity
(A) d
(C) d
Sol.
h
2
(B) d
3h
2
2h
(D) d
h
[AIEEE 2005]
on the surfaces of the earth and the moon respectively
and if Millikan's oil drop experiment could be performed
on the two surfaces , one will find the ratio
electronic ch arg e on the moon
to be [AIEEE 2007]
electronic ch arg e on the earth
(A) 1
(B) zero
(C)
gE
gM
Sol.
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(D)
gM
gE
GRAVITATION
Page # 53
Directions Question number 6 is Assertion-Reason type
question. This question contains two statements :
Statement I (A ssertion) an d Statemen t II
(Reason). The question also has four alternative
choices, only one of which is the correct answer. You
have to select the correct choice.
(A) Statement I is true; Statement II is true;
Statement II is not a correc t explanation for
Statement I.
(B) Statement I is true; Statement II is false.
(C) Statement I is false; Statement II is true.
(D) Statement I is true; Statement II is true;
Statement II is a correct explanation for Statement I.
18.The height at which the acceleration due to gravity
becomes
g
(where g = the acceleration due to gravity
9
on the surface of the earth) in terms of R, the radius
of the earth is
[AIEEE 2009]
R
(A) 2R
(B)
3
(C)
R
2
(D)
2R
Sol.
16.Statement I : For a mass M kept at the centre of
a cube of side a, the flux of gravitational field passing
through its sides is 4 GM.
and
Statement II : If the direction of a field due to a
point source is radial and its dependence on the
distance r from the source is given as
1
, its flux
r2
through a closed surface depends only on the strength
of the source enclosed by the
surface and not on
the size or shape of the surface.
[AIEEE 2008]
Sol.
19. Two bodies of masses m and 4 m are placed at a
distance r. The gravitational potential at a point on
the line joining them where the gravitational field is
zero, is
[AIEEE 2011]
17. A planet in a distant solar system is 10 times
more massive than the earth and its radius is 10 times
smaller. Given that the escape velocity from the earth
is 11 kms–1, the escape velocity from the surface of
the planet would be
[AIEEE 2008]
(A) 1.1 kms–1
(B) 11 kms–1
(C) 110 kms–1
(D) 0.11 kms–1
Sol.
(A)
4Gm
r
(B)
6Gm
r
(C)
9Gm
r
(D) zero
Sol.
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20. Two particles of equal mass m go around a circle
of radius R under action of their mutual gravitational
attraction. The speed of each particle with respect
to their centre of mass is
[AIEEE 2011]
(A)
Gm
R
(B)
Gm
4R
(C)
Gm
3R
(D)
Gm
2R
Sol.
Sol.
23.
F ou r p art icle s, e ac h of m as s M and
equidistant from each other, move along a circle of
radius R under the action of their mutual gravitiational
attraction. The speed of each particle is :
21. The mass of a spaceship is 1000kg. It is to be
launched from the earth's surface out into free space.
The value of g and R (radius of earth) are 10 m/s2
and 6400 Km respectively. The required energy for
this work will be
[AIEEE 2012]
(A) 6.4 x 1011 J
(B) 6.4 x 108 J
(C) 6.4 x 109J
(D) 6.4 x 1010J
Sol.
(A)
GM
1 2 2
R
(B)
(C)
GM
R
(D)
1 GM
1 2 2
2 R
2 2
GM
R
[JEE MAIN 2014]
Sol.
22. What is the minimum energy required to launch a
statellite of mass m from the surface of a planet of
mass M and radius R in a circular orbit at an altitude
of 2R?
[JEE MAIN 2013]
(A)
GmM
2R
(B)
GmM
3R
(C)
5GmM
6R
(D)
2 GmM
3R
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GRAVITATION
Page # 55
JEE ADVANCED
LEVEL - II
1. In a region of only gravitational field of mass ‘M’ a
particle is shifted from A to B via three different paths
in the figure. The work done in different paths are W1,
W2, W3 respectively then
B
(3)
C
M
(2)
(1)
(A) W1 = W2 = W3
(C) W1 = W2 > W3
Sol.
A
[JEE’ (Scr.) 2003]
(B) W1 > W2 > W3
(D) W1 < W2 < W3
3. A system of binary stars of masses mA and mB are
moving in circular orbits of radii rA and r B respectively.
If TA and TB are the time periods of masses mA and mB
respectively, then
[JEE 2006]
(A) TA > TB (if rA > rB)
(B) TA > TB (if mA > mB)
(C)
TA
TB
2
=
rA
rB
3
(D) TA = TB
Sol.
2. A body is projected vertically upwards from the
bottom of a crater of moon of depth R/100 where R is
the radius of moon with a velocity equal to the escape
velocity on the surface of moon. Calculate maximum
height attained by the body from the surface of the
moon.
[JEE’ 2003]
Sol.
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4. A spherically symmetric gravitational system of
particles has a mass density
[JEE 2008]
Sol.
R
0 for r
0 for r R
where 0 is a constant. A test mass can undergo circular motion under the influence of the gravitational
field of particles. Its speed V as a function of distance
r (0 < r < ) from the centre of the system is represented by
v
v
(A)
(B)
R
r
v
R
r
R
r
v
(C)
(D)
R
r
6. A thin uniform annular disc (see figure) of mass M
has outer radius 4 R and inner radius 3R. The work
required to take a unit mass from point P on its axis to
infinity is
[JEE 2010]
Sol.
P
4R
3R
4R
2GM
(4 2 – 5)
7R
GM
(C)
4R
Sol.
(A)
2GM
( 4 2 – 5)
7R
2GM
( 2 1)
(D)
5R
(B) –
5. STATEMENT-1
An astronaut in an orbiting space station above the
Earth experiences weightlessness.
[JEE 2008]
and
STATEMENT-2
An object moving around the Earth under the influence
of Earth’s gravitaitonal force is in a state of ‘free-fall’.
(A) STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True;
STATEMENT-2 is NOT a correct explanation for
STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True
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GRAVITATION
Page # 57
7. A binary star consists of two stars A (mass 2.2 Ms)
and B (mass 11 Ms), where Ms is the mass of the sun.
They are separated by distance d and are rotating
about their centre of mass, which is stationary. The
ratio of the total angular momentum of the binary star
to the angular momentum of star B about the centre
of mass is.
[JEE 2010]
Sol.
9. A satellite is moving with a constant speed 'V' in a
circular orbit about the earth. An object of mass 'm' is
ejected from the satellite such that it just escapes
from the gravitational pull of the earth. At the time of
its ejection, the kinetic energy of the object is
1
3
(A) mV2
(B) mV2
(C) mV2
(D) 2mV2
2
2
[JEE 2011]
Sol.
8. Gravitational acceleration on the surface of a planet
6
g. where g is the gravitational acceleration on
11
the surface of the earth. The average mass density
is
of the planet is
2
times that of the earth. If the
3
escape speed on the surface of the earth is taken to
be 11 kms–1, the escape speed on the surface of the
planet in kms–1 will be
[JEE 2010]
10. Two spherical planets P and Q have the same
uniform density , masses Mp and MQ, and surface
areas A and 4A, respectively. A spherical planet R
also has uniform density and its mass is (M P + MQ).
The escape velocities from the planets P, Q and R, are
Vp, VQ and VR, respectively. Then
[JEE 2012]
(A) VQ > VR > VP
(B) VR > VQ > VP
(C) VR / VP = 3
(D) VP / VQ
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1
2
GRAVITATION
Page # 58
Sol.
1
×(radius of Earth) has
10
the same mass density as Earth. Scientists dig a well
12.
A planet of radius R
R
on it and lower a wire of the same length
5
and of linear mass density 10-3 kgm-1 into it. If the
wire is not touching anywhere, the force applied at
the top of the wire by a person holding it in place is
(take the radi us oif E arth =6×1 0 6 m an d the
acceleration dur to gravity on Earth is 10 ms-2)
(A) 96 N
(B) 108 N
(C) 120 N
(D) 150 N
[JEE Main 2014]
Sol.
of depth
11. Two bodies, each of mass M, are kept fixed with a
separation 2L. A particle of mass m is projected from
the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G.
The correct statement(s) is (are)
[JEE 2013]
(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4
GM
.
L
(B) The minimum initial velocity of the mass m to esGM
.
L
(C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2
2GM
.
L
(D) The energy of the mass m remains constant.
Sol.
cape the gravitational field of the two bodies is
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GRAVITATION
Page # 59
OBJECTIVE PROBLEMS (JEE MAIN)
Exercise - I
1.
C
2.
D
3.
B
4.
B
5.
C
6.
B
7.
C
8.
A
9.
B
10.
B
11.
D
12.
A
13.
C
14.
A
15.
B
16.
B
17.
B
18.
C
19.
A
20.
C
21.
D
22.
D
23.
C
24.
B
25.
C
26.
B
27.
D
28.
C
29.
B
30.
A
31.
C
32.
B
33.
C
34.
A
35.
A
36.
B
37.
C
38.
C
39.
B
Exercise - II
1.
C
2.
A
3.
A
4.
B
5.
B
6.
B
7.
A
8.
A
9.
D
10.
B
11.
B
12.
C
13.
A
14.
C
15.
A, D
16.
A, D
17.
B,C,D
18.
A,C
19.
A,C
20.
B,D
21.
A
22.
A
23.
A,B,C
24.
A,D
(JEE ADVANCED)
Exercise - III
1. –
Gm 2
3Gm2
a
2.
3. (i)
2
3L
2 2 Gm
4 Gm
, (ii)
3 5R
3 R
4. 1.6 hours if is rotating from west to east, 24/17 hours if it is rotating from west to east.
5. 1 × 105 J
6.
2G
(sin ), (–G 2 )
R
Re k 2
9. (a) –GmMe /r, (b) –2GmMe/r
10.
3Gm m
R
3
G m
R 3
13. (i) –
15. g
17. 2
G 0R3
6
3
R / GM
M , (ii)
1
x–
R
2
2
–
7.
1 k
2
M
11. t
GMm 1 1
–
2C Re r
1
2
12.
8. h
–4GM2
3
L
5 –1
R
2
3
1
2
3
14. R
8
i , g – 2 G 0R i
x2
3
18. 1 –
2GM
1
a
16. (a)
Gm
; (b) r 2 ; (c)
r
2GM
r
x 2 – R2
4 R2
x
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GRAVITATION
Page # 60
1.
8.
2 R3 / 2 ( 6 6 )
GM (2 2
GM
R
3 3)
2
8
–
3
15
4 r 3/ 2
2.
G(4M m)
7
2
9. (a) h
3.
Gm
2d
4.
2
2
G R2 6.
3
1 R, (b) 1.13 R
Exercise - IV
10. T
6.4 km
sin–1
3
7. cos
1
3
10
Re
g
PREVIOUS YEAR QUESTIONS
JEE MAIN
LEVEL - I
1.
B
2.
A
3.
C
4.
D
5.
C
6.
C
7.
C
8.
A
9.
B
10.
A
11.
D
12.
D
13.
C
14.
C
15.
A
16.
A
17.
C
18.
A
19.
C
20.
B
21.
D
22.
C
23.
B
6.
12.
A
B
JEE ADVANCED
LEVEL - II
1.
7.
A
6
2.
8.
h = 99R
3
3.
9.
D
B
4.
10.
C
B, D
5.
11.
A
B
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CHEMICAL KINETICS
Page # 61
CHEMICAL KINETICS
This branch of chemistry which deals with the study of rates of chemical reactions and the mechanism
by which they occur. While studying reaction, one deals with :
(a) how fast (or slow) the reactants get converted into products
(b) the steps or paths through which the products are formed (reaction mechanism)
Chemical reaction kinetics deals with the rates of chemical processes. Any chemical process may be
broken down into a sequence of one or more single-step processes known either as elementary
processes, elementary reactions, or elementary steps. Elementary reactions usually involve either a
single reactive collision between two molecules, which we refer to as a a bimolecular step, or dissociation/isomerisation of a single reactant molecule, which we refer to as a unimolecular step. Very rarely,
under conditions of extremely high pressure, a termolecular step may occur, which involves simultaneous collision of three reactant molecules. An important point to recognise is that many reactions
that are written as a single reaction equation in actual fact consist of a series of elementary steps.
This will become extremely important as we learn more about the theory of chemical reaction rates.
As a general rule, elementary processes involve a transition between two atomic or molecular states
separated by a potential barrier. The potential barrier constitutes the activation energy of the process, and determines the rate at which it occurs. When the barrier is low, the thermal energy of the
reactants will generally be high enough to surmount the barrier and move over to products, and the
reaction will be fast. However, when the barrier is high, only a few reactants will have sufficient
energy, and the reaction will be much slower. The presence of a potential barrier to reaction is also the
source of the temperature dependence of reaction rates. The huge variety of chemical species, types
of reaction, and the accompanying potential energy surfaces involved means that the timescale over
which chemical reactions occur covers many orders of magnitude, from very slow reactions, such as
iron rusting, to extremely fast reactions, such as the electron transfer processes involved in many
biological systems or the combustion reactions occurring in flames. A study into the kinetics of a
chemical reaction is usually carried out with one or both of two main goals in mind:
1.
Analysis of the sequence of elementary steps giving rise to the overall reaction. i.e. the reaction
mechanism.
2.
Determination of the absolute rate of the reaction and/or its individual elementary steps.
RATE OF A REACTION
In general, for a reaction : R
P, the behaviour of the concentration of the reactant and product, as
the reaction proceeds is shown graphically
From the graph, it is clear that the concentration of the reactant decreases and that of the product
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CHEMICAL KINETICS
increases as the reaction proceeds and the rate of the change of the concentration of the reactant as
well as that of the product is also changing.
Rate of a reaction can, now, be defined in two ways :
Average Rate of reaction (rav) given by for the reaction R
rav = –
[R]
=
t
P:
[P]
t
where [R] and [P] represents the change in the concentrations of 'R' and 'P' respectively over a time
interval t
The average rate of the reaction between a time interval (tf – ti = t) can be determined from the
above graph by locating the concentration of 'R' of 'P' on this graph at the time instants tf and ti as
shown.
If [R]f and [R]i are the concentrations of the reactant 'A' at the time instants tf and ti then :
rav
–
[R]f – [R]i
tf – ti
Similarly from the plot of 'P' as a function of 't', we have : rav
[P]f – [P]i
tf – ti
Note :
The above expression for rav is equivalent to the slope of the line joining the points (tf , [A]f )
and ( ti , [R]i) or ( tf , [P]f ) and ( ti , [P]i) as shown.
Instantaneous Rate of reaction (rinst.) can be calculated from rav
in the limit t
0 and is represented as :
rinst. = –
Note :
d[R]
dt
d[P]
dt
The above expression for rinst. is equivalent to the slope of the tangent from the plot of the
concentration of 'R' or 'P' at any time instant 't'.
The rate of the reaction (rinst. or rav) is always calculated as a positive quantity.
The rate of a reaction at any temperature depends on the concentration of the reactants and
sometimes on the concentration of some foreign substances (e.g a catalyst) being used in the reaction
as well. The representation of this dependence of the rate of the reaction on the concentrations is
known as rate law and this rate law is determined experimentally.
Unit s
of
r ate
of
a
r eaction
–1
–1
Units of rate are concentration time . For example, if concentration is in mol L and time is in
–1 –1
seconds then the units will be mol L s . However, in gaseous reactions, when the concentration
of gas is expressed in terms of their partial pressures, then the units of the rate equation will be
–1
atm s .
Relation between various rates :
In general for a reaction : aA + bB
cC + dD
The rate of reaction can be expressed as follows :
Rate = –
1 d[A]
1 d[B]
1 d[C]
=–
=+
=
a dt
b dt
c dt
1 d[D]
= kr[A]m[B]n
d dt
ORDER OF A REACTION
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CHEMICAL KINETICS
Page # 63
By performing a reaction in actual in laboratory and carefully examining it, it is possible to express the
rate law as the product of concentrations of reactants each raised to some power. For example
consider the reaction : aA + bB
cC + dD. The differential rate law is written as :
Rate = –
1 d[A]
a dt
=–
1 d[B]
b dt
=+
1 d[C]
c dt
=
1 d[D]
d dt
= k r[A]m[B]n
where kr is called as rate constant of the reaction or velocity constant or specific reaction rate.
k is a characteristic of a reaction at a given temperature. It changing only when the temperature
changes.
The powers m and n are integers or fractions. m is called as order of reaction with respect to A and n
is called as order of reaction with respect to B.The overall order of reaction = m + n
Hence, the sum of powers of the concentration of the reactants in the rate law expression is
called the order of that chemical reaction.
The values of m and n are calculated from the experimental data obtained for a reaction and
the powers m and n are not related to the stoichiometric coefficients of the reactants
Order can be fractional, zero or negative.
For example consider the following reaction :
(i)
H2(g) + Br2(g)
(ii)
CH3CHO(g)
2 HBr (g) rate = k[H2] [Br2]1/2 (by experiment), order of reaction = 1 + 1/2 = 3/2
CH4(g) + CO(g), rate = k[CH3 CHO]3/2 , order of reaction = 3/2
Units of k :
dc
dt
where k : rate constant; c : concentration and n : order of reaction
–kc n
In general, the rate law for a nth order reaction can be taken as :
k
dc / dt
cn
Units of k
(mol/L)1–n (time)–1
For a 'zero' order reaction (n = 0) : Units of k = (mol/L)1 (time)–1 or mol/L/sec
For a first order reaction (n = 1) : Units of k
(time)–1 e.g. sec–1, min–1, hrs–1 etc.
For a second order reaction (n = 2) : Units of k
(mol/L)–1 (time)–1 or L/mol/sec.
MOLECULARITY
As already discussed, the order of a reaction is an experimental concept.
A complex chemical reaction is understood in terms of various indirect steps called elementary processes.
The study of a reaction in terms of elementary processes is called as reaction mechanism. Now various
elementary steps occur at different rates.
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction,
which must collide simultaneously in order to bring about a chemical reaction is called
molecularity of a reaction.
In the rate determining step, when one molecule takes part, it is said to be a unimolecular reaction ;
two molecules take part, it is said to be a bimolecular reaction; three molecules take part, it is said to
be a termolecular reaction.
Unimolecular :
1.
Cyclopropane propene
2.
O3(g) O2(g) + O(g)
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CHEMICAL KINETICS
3.
N2O5(g)
N2O4(g) + 1/2O2(g)
Bimolecular :
1.
NO(g) + O3 (g) NO2(g) + O2(g)
2.
2HI(g) H2(g) + I2(g)
Termolecular :
1.
2NO(g) + O2(g)
2NO2 (g)
The probability that more than three molecules can collide and react simultaneously is very small.
Hence, reactions with the molecularity three are very rare and slow to proceed. It is, therefore,
evident that complex reactions involving more than three molecules in the stoichiometric equation
must take place in more than one step.
KClO3 + 6FeSO4 + 3H2 SO4
KCl + 3Fe 2(SO4)3 + 3H2O
This reaction which apparently seems to be of tenth order is actually a second order reaction. This
shows that this reaction takes place in several steps. Which step controls the rate of the overall
reaction? The question can be answered if we go through the mechanism of reaction, for example,
chances to win the relay race competition by a team depend upon the slowest person in the team.
Similarly, the overall rate of the reaction is controlled by the slowest step in a reaction called the rate
determining step.
(i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but
molecularity cannot be zero or a non integer.
(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions. For complex reaction molecularity has no meaning. 103
Chemical Kinetics
(iii) For complex reaction, order is given by the slowest step and molecularity of the slowest step
is same as the order of the overall reaction.
For a reaction : A
B
in the rate law : rate = k[A]m [B]n
Neither the order of reaction (m + n) nor the molecularity of a reaction can be predicted from
stoichiometric coefficient of a balanced reaction. The order of reaction is always to be determined
experimentally and molecularity is determined theoretically after studying the reaction mechanism.
However as a theoretical idea sometime, we can have an approximate order of reaction equal to
molecularity (i.e., the number of molecules taking part in slowest elementary for complex reactions).
Problem 1 :
The rate of formation of NO(g) in the reaction NOBr(g) NO(g) + Br2(g) is found to be 1.6 × 10–4 M/s.
Find the rate of overall reaction rate and rate of consumption of NOBr.
d[NO]
1.6 × 10–4 M/s.
dt
First write a balanced chemical equation.
We have :
Now, Rate of overall reaction = –
Rate of consumption of NOBr = –
2NOBr(g)
1 d[NOBr ]
=
2
dt
2NO(g) + Br2(g)
1 d[NO]
1 d[Br2 ]
=
= 0.8 × 10–4 M/s
2 dt
1 dt
d[NOBr ]
= + 1.6 × 10–4 M/s
dt
Problem 2 :
The rate constant for a given reaction is k = 3 × 10–5 s–1 atm–1. Express it in units of L mol–1 sec–1.
Sol.
PV = nRT
P = cRT (c : concentration in mol/L)
Substitute R = 0.0821 L–atm/mol/K ; T = 273 K ; P = 1 atm
c = 0.04462 mol/L
k
3 10 –5
= 6.73 × 10–4 L/mol/s.
0.04462
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CHEMICAL KINETICS
Page # 65
Problem 3 :
From the rate laws for the reactions given below, determine the order with respect to each
species and the overall order :
(i)
2HCrO4– + 6I– + 14H+
2Cr3+ + 3I2 + 8H2O,
Rate = k[HCrO4–] [I–]2 [H+]2
–
+
(ii)
H2O2 + 2I + 2H
I2 + 2H2O,
Rate = k[H2O2] [I–]
Sol.
(i)
The order of the reaction with respect to [HCrO4–] is 1; with respect to [I–] is 2 and with
respect to [H+] is 2. The overall order of the reaction is 1 + 2 + 2 = 5
(ii)
The order of the reaction with respect to [H2O2] is 1 and with respect to [I–] is 1. The overall
order of the reaction is 1 + 1= 2.
In (i) stoichiometric coefficient of I– is 6 whereas the power coefficient (n) in the rate law is 2.
Reaction (i) may not take place in a single step. It may not be possible for all the 22 molecules
to be in a state to collide with each other simultaneously. Such a reaction is called a complex reaction.
A complex reaction takes place in a series of a number of elementary reactions.
Zero Order Reactions :
The rate law for zero order reactions (n = 0) is written as :
A
product
t=o
a = [A]o
o
t=t
a – x = [A]
x
[A]
[A]0
Slope = –k
d [A]
–
dt
–
d [A]
dt
k [A]
t
k
[A]
–
t
d [A]
k
[A] o
dt
o
[A]o – [A] = kt
k
[A] o – [A]
x
=
t
t
Half life (t 1/2) :
Time in which half of initial amount is left.
[A]o
[A]o/2
t=o
k=
t 1/ 2
t = t1/2
[A]o – [A]o /2
t1/2
t1/2 =
A0
[A]o
2k
Thus, for a Zero order reaction, half life is directly proportional to initial concentration of the reactant.
Clearly, zero order reactions are those, whose rates are not affected by change in concentrations of
reactants (i.e., independent of concentration). The rates of such reactions only depend upon
temperature. Most of photochemical reactions are zero order reactions. Other examples are :
decomposition of HI over the surface of gold and NH3 over tungsten.
Example :
(1) Photochemical Reactions, Photosynthesis
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CHEMICAL KINETICS
(2)
CH4 + Cl 2
h
Cl – Cl
Cl
CH4 + Cl
CH3Cl + H
FIRST ORDER REACTION
A
product
t=o
[A]o
–
t=t
[A]
[P]
–
d [A]
dt
Rate
[A]0
[A]0
k [A]
t
[A]
–
[A]
t
t
d[A]/[A]
k
[A]o
dt
o
ln [A]o/[A] = kt
k
2.303
[A]o
log
t
[A]
Half life (t1/2) :
t 1/ 2
t = t1/2 [A] = [A]o/2
k
2.303
[A]o
log
t1/2
[A]o /2
t1 / 2
t1 / 2
[A] =
=
2.303
log 2
t 1/ 2
A
2.303
log10 2
k
0.693
k
[ A]o
2n
t1 / 2
loge 2
k
where , n = number of half lifes.
Average life :
tav = 1.44 t1/2 =
1.
2.
3.
1
k
Features of a First Order Reaction :
A first order reaction must follow above form of rate law for all time instants.
This means if we are given value of A0 and values of x at different time instants [i.e.(A0 – x) as value
of reactants after t], the values of k can be calculated for different time instants by using the above
first order law.
If the reaction for which the data were given is a first order reaction, then all values of k will
approximately equal to each other.
The time for half reaction for a first order reaction is independent of initial concentration of
reactants.
The concentration of reactants in a first order reaction decreases exponentially with time (see figure)
Note that plot of log10 A vs t is linear.
Example :
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CHEMICAL KINETICS
(1)
Page # 67
log10 [A]
Radioactive disintegration.
(2) PCl 5(g)
(3) H 2O2
log [A]0
PCl 3(g) + Cl2(g)
Slope
H2O + ½O2
(4) NH4NO2
k
2303
.
N2 + 2H2O
t
Rate constant of a first order reaction can also be calculated by measuring the concentration
of the reactants at two time instants (if the initial concentration is not known).
If A1 and A2 are the reactant's concentrations at two time instants 't1' and 't2' respectively, then we
have :
2. 303log10
A0
A1
kt1
...(iii)
A0
and 2.303log10 A
2
kt 2
... (iv)
Subtracting (iv) from (iii), we get :
2. 303log10
A1
A2
k( t 1 t 2 )
Thus, k can be evaluated.
Problem : 4
For a reaction 2NO(g) + 2H2(g) N2(g) + 2H2O (g) ; the following data were obtained.
[NO] (mol/L)
[H2](mol/L)
Rate (mol/L/s)
1.
5 × 10–3
2.5 × 10–3
3 × 10–5
–3
–3
2.
15 × 10
2.5 × 10
9 × 10–5
3.
15 × 10–3
10 × 10–3
3.6 × 10–4
(a) Calculate the order of reaction.
(b) Find the rate constant.
(c) Find the initial rate if [NO] = [H2] = 8.0 × 10–3 M
Assuming rate law can be expressed as follows :
rate = k[NO]x [H2]y
By analysing the data :
From observation 1 and 2, we see that [H2] is constant and when [NO] is tripled, the rate is also
tripled.
rate (r) [NO]
x=1
From observations 2 and 3, we see that [NO] is constant; when [H2] is increased four times, the rate
also increases four times :
rate
[H2]
y=1
r = k [NO] [H2O]
The order of reaction w.r.t No and H2 is 1 and the overall order of reaction is 1 + 1 = 2.
Initial rate = k[NO][H2] = 2.4 × (8 × 10–3)2 = 1.536 × 10–4 mol/L/s.
Second Order kinetics
Case I :
A
product
t=o
[A]o
–
t = t
[A]t
[P]
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CHEMICAL KINETICS
d [A]
= k[A]2
dt
–
[A]
d[ A]
[A]o
[ A] 2
–
1
[A]
k=
–
Slope=k
t
k
1
[A] t
dt
o
Time
1
[A]o
kt
1/[A] – 1 /[A] o
t
Half-life (t1/2) :
1
[A] 0
[Ao ]
, k = 2/[A]o – 1/[A]o
2
[A]t =
t1/2 =
t1 /2
Slo pe=k
1
[A]0
1
k [A]o
Example :
(1) Alkaline hydrolysis of esters.
(2) Self Canizzaro’s reactions,
Case (II):
A
+
t=o
a
t=t
a–x
d [A]
dt
–
x
o
B
b
–
b–x
dx
(a – x) (b – x)
x
dx
k (a – x) (b – x)
dt
k[ A] [B] ,
t
x
o
o
k dt ,
(–1)
a–x
ln
(b – a)
a
ln
product
(–1)
b –x
ln
(a – b)
b
dx
(b – a) (a – x)
x
o
t
dx
k dt
o
(a – b ) (b – x)
kt
a
b
– ln
k (b – a) t
a–x
b–x
k=
1
t (b – a)
k
1
a (b – x)
ln
t (b – a)
b (a – x)
ln
a
a–x
b–x
b
Pseudo First Order Reaction
The order of a reaction is sometimes altered by conditions. Consider a chemical reaction between
two substances when one reactant is present in large excess. During the hydrolysis of 0.01 mol of
ethyl acetate with 10 mol of water, amounts of the various constituents at the beginning ( = 0)
and completion (t) of the reaction are given as under.
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CH3COOC2H5 + H2O
CH3COOH + C2H5OH
H
t=0
0.01 mol
10 mol
0 mol
0 mol
t=t
0 mol
9.9 mol
0.01 mol
0.01 mol
The concentration of water does not get altered much during the course of the reaction. So, in the
rate equation Rate = k'[CH3COOC2H5] [H2O] the term [H2O] can be taken as constant. he equation, thus, becomes Rate = k[CH3COOC2H5] where k = k [H2O] and the reaction behaves as first
order reaction. The molecularity of acidic hydrolysis of sucrose and esters is 2, whereas their order
is 1. In both the reactions water is in excess so that its concentration remains constant throughout
the reaction. The rate of reaction therefore depends only on the concentration of sucrose and ester
in two reactions respectively. So the reactions in which the molecularity is 2 or 3 but they conform
to the first order kinetics are known as pseudo first order reactions OR pseudo unimolecular reactions.
C12H22O11 + H2O + H+ C6H12O6(glucose) + C6H12O6(fructose)
CH3COOC2H5 (ester) + H2O + H+
CH3COOH + C2H5OH
(In both the reactions, H ion acts as a catalyst)
+
nTH ORDER KINETICS
A
product
t1/2
n -1
[A]t
1
[A]0n–1
t
k dt
n
[A]
[A]o
n
k[A]
d [A]
–
2 –1
(n-1)k
Slope=
d [A]
–
dt
o
1
1
(n – 1) [A]n–1
1
[A]n–1
t
[A ]
kt
Slope=k(n–1)
1
[A]0n–1
[A ]o
Time
1
1
1
–
(n – 1) [A]t n – 1 [A]o n –1
k
1
(n – 1) t
Half-life (t1/2) :
1
k = (n – 1) t
1/2
1
1
–
n –1
[A]t
[A]o n–1
at t = t1/2 , [A]t =
2n –1
[A]o
n –1
kt
–
[ A]o
2
1
[A]o n –1
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CHEMICAL KINETICS
t 1/2 =
2n–1 – 1
(n –1) k [A]o n–1
t1 / 2
1
[A]o n –1
Problem : 5
For the non-equilibrium process, A + B Products, the rate is first order w.r.t A and second
order w.r.t. B. If 1.0 mole each of A and B are introduced into a 1 litre vessel and the initial rate
were 1.0 × 10–2 mol/litre-sec, calculate the rate when half of the reactants have been used.
Sol.
Rate = K[A] [B]2
10–2 = K[1] [1]2
or
K = 10–2 litre2 mol –2 sec –1
Now
ratell = 10–2 × 0.5 × (0.5)2
or
New rate = 1.2 × 10–3 mol/L-sec
ANALYSIS OF SOME IMPORTANT FIRST-ORDER REACTIONS
Decomposition of Hydrogen peroxide (H2O2)
H2O2(g)
H2O(g) + ½ O2(g)
The rate of this first order reaction is measured by titrating a fixed volume of H2O2 (undecomposed)
against a standard solution of KMnO4. Here KMnO4 acts as oxidising agent and H2O2 as reducing agent.
The volumes of KMnO4 used for H2O2 after regular intervals of time are as follows.
Time instants
t=0
t1
t2
t3
t4
t5
Vol. of KMnO 4
V0
V1
V2
V3
V4
V5
Volume of KMnO4 at t = 0 corresponds to volume of H2O2 initially present.
A0
V0
Volume of KMnO4 at time instants t1, t2, t3 , .................... corresponds to volume of H2O2
re mai nin g
after
t1, t2, t3 , .................
A
Vt
Now it being a first order reaction, follows first order kinetics, so
V0
k t = 2.303 log10 V
t
Now using the above expression, if we calculate the values of k for different time intervals t1, t2,
........... (for actual numerical data), the values of k should be same if the reaction follows first order
kinetics.
Decomposition of ammonium nitrite (NH4NO2)
and benzene diazonium chloride (C6H5N = NCl)
NH4NO2(g)
2 H2O(g) + N2(g)
C6H5 – N = N – Cl(g) C6H 5 – Cl(g) + N2(g)
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The rate of both the reaction is studied (measured) in similar manner. The volume of nitrogen (N2) is
collected after a regular interval of time as follows :
Tim e instants
t = 0
t1
t2
t3
t4
Vol. of N 2
0
V1
V2
V3
V4
t
V
At t = 0, clearly the volume of N2 = 0
Time instant t = means the end of a reaction i.e., when whole of NH4NO2 or C6H5 – N = N – Cl is
decomposed.
A t t = , V corresponds to the initial volume of NH4NO2 or C6H5 – N = N – Cl
(Note that the ratio of stoichiometric coefficient for both N2 : NH4NO2 or N2 : C6H5N = NCl is 1 : 1)
A0
V
A t t = t1, t2, t3................ the volume of N2 corresponds to concentration of product formed i.e., equal
to x.
x Vt
A0 – x
V – Vt
V
k t = 2.303 log10 V – V
t
Hydrolysis of Esters (CH3COOC2H5)
CH3COOC2H5 (ester) + H2O + HCl(H+) CH3COOH + C2H5OH
The reaction rate is measured by titrating the acid (CH3COOH) produced against a standard alkali
solution. Note that when a test sample is prepared from the reacting mixture, there are two acids :
one is minral acid H+ (HCl or any other) and second is CH3COOH produced. So volume of alkali used
gives titration value for both acids. The data is collected in the following manner.
Time instants
t=0
t1
t2
t3
t4
t
Vol. of NaOH
V0
V1
V2
V3
V4
V
At t = 0, V0 is the volume NaOH used to neutralise the mineral acid present (H+) being used as catalyst.
(At t = 0, no CH3COOH is yet produced)
At t = (i.e., at the end of hydrolysis), V , is the volume of NaOH used to neutralise whole of CH3COOH
plus vol. of HCl present At t = , volume of CH3COOH corresponds to volume of ester taken initially
A0 V
– V0
(as V0
vol. of HCl)
At t = t1, t2, t3 ............ V1, V2 , V3, ................corresponds to vol. of HCl plus vol. of CH3COOH being
produced.
x V t – V0
A0 – x (V – V0 ) – (Vt – V0 )
A0 – x
V – Vt
k t = 2.303 log10
V – V0
V – Vt
Inversion of Cane Sugar (C12H22O11)
C12H22O11 + H2O + H+ C6H12O6(glucose) + C 6H12O6 (fructose)
The rate is measured by measuring the change in the angle of rotation (optical activity) by a polarimeter.
Sucrose is dextro-rotatory, glucose is dextro-rotatory and fructose is leavo-rotatory. The change
produced in rotatory power in time t gives a measure of x, the quantity of sucrose decomposed in that
time. The total change in the rotatory power produced at the end of the reaction gives the measure of
A0, the initial concentration of sucrose.
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CHEMICAL KINETICS
If r0, r1 and r represent rotations at the start of reaction, after time t and at the end of reaction
respectively,
then
A0 r0 – r
and
x r0 – rt
A0 – x rt – r
r0 – r
kt = 2.303 log10 r – r
t
DECOMPOSITION OF AsH3(g)
In first-order reactions involving gases, sometime measuring the pressure of the reaction mixture is
very good method for measuring reaction rates.
For example consider decomposition of arsine gas (AsH3)
3
H2 ( g)
2
The rate of reaction is measured as the increase in pressure of the reaction mixture. Note that there is
an increase in number of moles of the gaseous products to the right, so as the reaction proceeds,
there will be an increase in pressure of contents (P n).
Let the initial pressure of AsH3(g) is P0, if x is the decrease in pressure of AsH3(g) after time t.
AsH3(g)
As(s)
+
H2 (g)
A0 initial pressure
P0
0
0
AsH3(g)
A
As(s) +
partial pressures
P0 – x
0
3
x
2
Arsenic is solid, so P(AS) = 0
After time t, let Pt be the total pressure, then
Pt = P(AsH3) + P(H 2) = (P0 – x) +
Pt = P0 +
1
x
2
Now A0 P0
and A P0 – x
3
x
2
x = 2(Pt – P0)
P0 – 2 (Pt – P0)
3P0 – 2Pt
P0
k t = 2.303 log10 3P – 2P
0
t
On similar pattern, please try to write the expression for Ist order rate law for following first-order
reactions. (in terms of P0 and Pt )
1. N2O(g)
N 2(g) +
1
O (g)
2 2
2. (CH3)3C – O – O – C(CH3)(g)
2(CH3)2C = O(g) + C2H6(g)
COMPLEX (FIRST ORDER) KINETICS
(A) Parallel Kinetics
K1
A
B
I
K2
II
C
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Rate of change of A = [rate of change of A]I + [rate of change of A]II
–
d [A]
dB
= K1 [A] + K2 [A],
dt
dt
k1A ,
k1
% of B in the mix of A & B = k k
1
2
% of C in the =
d [A]
(K1
dt
K1
k2 A
k1
k2
[B]
[ C]
–
dC
dt
k2
k1 k 2
100
100
K2 ) [A]
2.303
[A]o
log10
t
[A]t
K2
0.693
K1 K2
t1 / 2
GENERALIZATION
k1
B
C
k2
D
A
K1
kn
K2
- - - - - - Kn
2.303
[A]
log o ,
t
[ A]t
t1 / 2
K1
0.693
K2 - - - - Kn
Z
Problem : 6
An organic compound A de composes following two parallel first order mechanisms :
k1
B
k1 1
–5
–1
A
; k
9 and k1 = 1.3 × 10 sec .
k2
2
C
Calculate the concentration ratio of C to A, if an experiment is allowed to start with only A for
one hour.
Sol.
k1 1
k2 9
But k1 = 1.3 × 10–5 sec–1 ; k 2 = 9 × 1.3 × 10–5 sec–1 = 117 × 10–5 sec–1
(k 1 + k2) = (1.3 × 10–5 ) + (11.7 × 10–5) sec–1 = 13 × 10–5 sec–1
....(1)
[B] t
Also [C ]
t
ln
[ A]0
[ A] t
1
9
[B]t =
[C] t
9
(k1 k 2 ) t ; ln
[ A] t
...(2)
[B] t
[ A] t
[ C] t
= (k1 + k 2)t
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ln
CHEMICAL KINETICS
[ A] t
[C] t
9
[ A] t
10 [ C] t
9 [ A] t
ln 1
[C] t
[from eq. (2)]
(k1 k2 ) t
(k1 k 2 ) t
= 13 × 10–5 × 60 × 60 = 0.468 [from eq. (1)]
[C] t
10 [ C] t
= 1.5968 ; [ A]
[
A
]
9
t
t
1+
10 [C] t
9 [ A] t
ln 1
0.537
(2) SERIES KINETICS
A
–
d [A]
dt
K1
B
K2
C
K1 [A],
[ A]o e –k1t
[ A]t
d [B]
K1 [A] – K2 [B],
dt
[B]
k1 [A]o
(k2 – k1 )
d [C]
dt
[C ]
k1 k2 [A]o
k1 – k2
K2 [B],
e –k1t – e –k 2 t
(e–k1t – 1) (e –k2 t – 1)
–
k1
k2
Graph of [A], [B], [C] Vs t:
[A]
[C]
[A]0
[A]0
t
t
Time when [B] is maximum
[B]
k1 [A]o
e – k1t – e –k2 t
[B] =
k2 – k1
d [B]
dt
k1 e –k1t
k1 [A]o
– k1e – k1t – (–k2 ) e –k2 t
k2 – k1
0
t
k 2e –k 2t
t
k
loge 2
k1
( k2 – k1 )
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e
k2
k1
(k2 – k1) t
(k2 – k1)t
t=
Page # 75
log e
k2
k1
log e (k2 /k1)
k2 – k1
REVERSIBLE KINETICS
Conc.
A
B
a
t=0
a
0
t = t1
a – x1
x1
t = t2
a – x2
x2
t = t3
a – x3
x3
- - - -- - - - - -- - -- - - - - - - - -- - - - - - - -- - - - t = teq a – x
x
O < t < teq
rf =
[A]o – [A] equilibrium
[ A]equilibriu m
[ A]equilibrium
rb ,
[B]
a–x e
[A]
t1
CHEMICAL KINETICS, teq
At equilibrium,
k1
k2
A
xe
t2
t3
teq
t <
CHEMICAL EQUILIBRIUM
k1
k1 [A] equilibrium = k2 [B] equilibrium , k
2
k1 k 2
[A]o
[A]eq – 1, k 2
time
[B]eq.
[A]eq.
[A]o
[A]eq
k 2 [A]o
k1 k2 ............ (I)
d [A]
d [A]
= – k1 [A] + k 2 [B],
= – k1 [A] + k2 [ [A]o – [A] ] = – (k1 + k2 ) [A] + k2 [A]o
dt
dt
= (k1 + k 2)
– [A]
k2
[A]o
k1 k2
By substituting the value from equation (I)
d [A]
dt
log e
(k1
[A]t
(k1 k 2 ) [A]eq – [A]
[A]eq – [A]o
[ A]eq – [ A] t
k2 )
[A]o
d [A]
[A]eq – [A]
t
(k1 k2 )
dt
o
(k1 k2 ) t
[A]eq – [A]o
2.303
log 10
t
[ A]eq – [A]t
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CHEMICAL KINETICS
Problem : 7 For a reversible first order reaction,
Kf
A
B;
Kf = 10–2 sec –1
Kb
Beq
and A = 4 ; If A0 = 0.01 ML–1 and B0 = 0, what will be concentration of B after 30 sec ?
eq
Sol.
A0
0.01
0.01 – xeq
[B] eq
=
[ A] eq
B
0
xeq
[ x] eq
4 =
0.01 – [ x] eq
10 –2
Kb
Kb = 0.25 × 10–2
and
xeq =
0.04
= 0.008
5
[ x] eq
2.303
log
t = (K
K
)
[
x
] eq – x
f
b
30 =
2.303
log
10 –2
125
.
0.008
0.008 – x
0.008
( 0.008 – x)
1455
.
x = 2.50 × 10–3
Ways to determine order of reaction
Half life method
Initial Rate
method
Integrated Rate law
method
Ostwald Isolation
method
(1) Initial Rate Method
A+B
product
rate = k [A]m [B]n
;
Order = m + n
[A]
[B]
rate
Experiment 1
0.1
0.1
2 × 10–3
Experiment 2
0.1
0.2
4 × 10–3
Experiment 3
0.2
0.1
32 × 10–3
Experiment (1) and Experiment (2)
2 10 –3
4 10 – 3
k [0. 1]m [0.1]n
k [0.1]m [0.2]n
1
1
1
2
2
n= 1
n
Experiment (1) and Experiment (3)
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2 10 –3
32 10 –3
1
2
4
k [0.1]m [0.1]n
k [0.2]m [0.1] n
m
1
2
m=4
Order (m + n) = 4 + 1 = 5
(2) Half - life method
t1/2
[A]o
1 hr
0.1
2 hr
t1/2
1
[ A]n –1
0.2
[t 1/ 2 ]1
[ t1 / 2 ]2
1
2
1
2
–1
[0.1]n –1
[0.2]n –1
–1
1
2
n –1
n=0
n
1
log (t1/2 )2 / (t1/2 )1
log [ A]o1 /[ A]o 2
(3) Integrated Rate law Method
A
t=0
product
1000 M
t = 60 sec
t = 120 sec
100 M
10 M
[ A] o – [A]t
t
1000 – 10
990
k =
120
120
n = 1
n= 0 k=
1000 – 100
900
=
= 15
60
60
k=
2.303
1000
log
60
10
2.303
60
k=
2.303
1000
log
120
10
2.303
60
(4) Ostwald Isolation method
rate = k[A]m [B]n [C]o [D]p - - - - - - - Experiment 1 : [A] = In small quantity ; [B], [C], [D] - - - - - - - in excess
The rate equation reduces to
rate = k’ [A]m
r 1 = k’ [A]1m
r2 = k’ [A]2m
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CHEMICAL KINETICS
r1
r2
[A]1
[A]2
m
, log
r1
r2
m log
[ A]1
[ A]2
log (r1 /r2 )
log ([A]1 /[A]2 )
m
Experiment 2: [B] = In small quantity . & [A], [C], [D] - - - - - rate = k’ [B]
n
in excess.
repeated
Order of reaction = m + n + o + p + - - - - - - - ACTIVATION ENERGY (Ea)
A mixture of magnesium and oxygen does not react at room temperature. But if a burning splinter is
introduced to the mixture, it burns vigorously. Similarly a mixture of methane and oxygen does not
react at room temperature, but if a burning match-stick is put in the mixture, it burns rapidly. Why it
happen like this, that some external agents has to be introduced in order to initiate the reaction ?
According to the theory of reaction rates "for a chemical reaction to take place, reactant molecules
must make collisions among themselves". Now in actual, only a fraction of collisions are responsible for
the formation of products, i.e., not all collisions are effective enough to give products. So the collisions
among reactant molecules are divided into two categories :Effective collisions and In-effective collisions
Effective collisions are collisions between the molecules which have energies equal to or above a
certain minimum value. This minimum energy which must be possessed by the molecules in order to
make an effective collision (i.e., to give a molecule of products) is called as threshold energy. So it is
the effective collisions which bring about the occurrence of a chemical reaction.
Ineffective collisions are the collisions between the molecules which does not posses the threshold
energy. These can not result in a chemical reaction.
Now most of the times, the molecules of reactants do not possess the threshold energy. So in order to
make effective collisions (i.e., to bring about the chemical reaction), an additional energy is needed to
be absorbed by the reactant molecules. This additional energy which is absorbed by the molecules so
that they achieve the threshold energy is called as energy of activation or simply activation energy. It
is represented as Ea.
A reaction which needs higher activation energy is slow at a given temperature.
E
E
Ea
Ea
ER
EP
H 0
H 0
EP
ER
time
time
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For example : NO(g) +
1
O (g)
2 2
NO2(g) is faster at ordinary temperature whereas the following
reaction :
1
O 2 ( g)
2
is much higher.
CO(g) +
CO2(g) is slower at the same temperature as the value of Ea for the second reaction
FACTORS AFFECTING RATE OF REACTION
(a)
Catalyst : The rate of reaction increased by addition of catalyst, because catalyst lowers, the
activation energy and increased the rate of reaction. A catalyst is a substance which increases the
rate of a reaction without itself undergoing any permanent chemical change. For example, MnO2
catalyses the following reaction so as to increase its rate considerably.
2KClO3
MnO2
2 KCl + 3O2
The word catalyst should not be used when the added substance reduces the rate of raction. The
substance is then called inhibitor. The action of the catalyst can be explained by intermediate complex
theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary
bonds with the reactants resulting in an intermediate complex. This has a transitory existence and
decomposes to yield products and the catalyst.
It is believed that the catalyst provides an alternate pathway or reaction mechanism by reducing the
activation energy between reactants and products and hence lowering the potential energy barrier as
shown in Fig. It is clear from Arrhenius equation that lower the value of activation energy faster will be
the rate of a reaction. A small amount of the catalyst can catalyse a large amount of reactants. A
catalyst does not alter Gibbs energy, G of a reaction. It catalyses the spontaneous reactions but
does not catalyse nonspontaneous reactions. It is also found that a catalyst does not change the
equilibrium constant of a reaction rather, it helps in attaining the equilibrium faster, that is, it catalyses
the forward as well as the backward reactions to the same extent so that the equilibrium state remains
same but is reached earlier.
(b)
Temperature :
With increase in temperature the rate of reaction increases. It is generaly found
o
for every 10 increase in temperature. The rate constant double.
The ratio of rate constants with 10o difference in their temperature is called temperature coefficient.
KT 10
= Q = Temperature coefficient of reaction
KT
(c)
2
Concentration :
Rate = A e–Ea/RT [A] m [B] n - - - - - - - - With increase in concentration of reactants the rate of the reaction increases because number of
collision (effective collisions) increases.
(d)
Nature of Reactants :
Ionic Reactants :
Generaly ionic reactions in aq. media are fast than the reaction involving covalent reactants.
As covalent reactants involving breaking of bond then formation of bond where as ionic reaction
involve in single step.
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(e)
Surface Area : Increase in surface area increases the number of collisions and hence rate increases
(f)
Radiation : Some reactions exposes to sunlight also increases the rate of reaction.
Collision Theory of Chemical Reactions
Though Arrhenius equation is applicable under a wide range of circumstances, collision theory, which
was developed by Max Trautz and William Lewis in 1916 -18, provides a greater insight into the
energetic and mechanistic aspects of reactions. It is based on kinetic theory of gases. According to
this theory, the reactant molecules are assumed to be hard spheres and reaction is postulated to
occur when molecules collide with each other. The number of collisions per second per unit volume of
the reaction mixture is known as collision frequency (Z). Another factor which affects the rate of
chemical reactions is activation energy (as we have already studied). For a bimolecular elementary
reaction
A+B
Products rate of reaction can be expressed as
Rate = ZAB e
Ea / RT
..............(1)
where ZAB represents the collision frequency of reactants, A and B and e Ea / RT represents the fraction
of molecules with energies equal to or greater than Ea. Comparing (1) with Arrhenius equation, we can
say that is related to collision frequency. Equation (1) predicts the value of rate constants fairly
accurately for the reactions that involve atomic species or simple molecules but for complex molecules
significant deviations are observed. The reason could be that all collisions do not lead to the formation
of products. The collisions in which molecules collide with sufficient kinetic energy (called threshold
energy*) and proper orientation, so as to facilitate breaking of bonds between reacting species and
formation of new bonds to form products are called as effective collisions.
To account for effective collisions, another factor P, called the probab ilit y or s teri c factor is
introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e.,
Rate = PZ AB e Ea / RT
Thus, in collision theory activation energy and proper orientation of
th e mol ec ul es toget her
determine the criteria for an effective collision and hence the rate of a chemical reaction.
MECHANISM OF A REACTION
Reactions can be divided into
Elementary/Simple/single step
Complex/multi-step
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ELEMENTARY REACTION :
These reaction take place in single step without formation of any intermediate
T.S.
Ep
Er
For elementary reaction we can define molecularity of the reaction which is equal to no of molecules
which make transition state or activated complex because of collisions in proper orientation and with
sufficient energy
molecularity will always be a natural no
1 = unimolecular one molecular gets excited (like radioactivity)
2 = bimolecular
3 = trimolecular
Molecularity < 3 because the probability of simultaneous collision between 4 or mor molecules in proper
orientation is very low
For elementary reaction there is only single step and hence it is going to be rate determining step so
order of an elementary reaction is its molecularity
Order of elementary reaction w.r.t reactant = stoichiometric co-efficient of the reactant
H2
I2
2HI
Simple reaction
rate = k [H2] [I2]
2H2 2I2
4HI (no elementary)
reaction obtained by multiplying and elementary reaction with some no will not be of elementary nature
H2 Cl2
2HCl
order = 0
COMPLEX REACTION :
Reaction which proceed in more than two steps or having some mechanism. (sequence of elementary
reaction in which any complex reaction proceeds)
For complex reaction each step of mechanism will be having its own molecularity but molecularity of
net complex reaction will not be defined.
Order of complex reaction can be zero fractions whole no, even negative w.r.t some species.
Order of reaction or rate law of reaction is calculated with the help of mechanism of the reaction
generally using rate determine step (R.D.S) if given.
Rate law of a reaction is always written in terms of conc. of reactant, products or catalysts but never
in terms of conc. of intermediates.
The mechanism of any complex reaction is always written in terms of elementary steps, so molecularity
of each of these steps will be defined but net molecularity of complex reaction has no meaning.
The mechanism of most of the reaction will be calculated or predicted by using mainly the following
approximation.
The Rate-Determining-Step Approximation %
In the rate-determining-step approximation (also called the rate-limiting-step approximation or the
equilibrium approximation), the reaction mechanism is assumed to consist of one or more reversible
reactions step, which in turn is followed by one or more rapid reactions. In special cases, there may be
no equilibrium steps before the rate-determining step or no rapid reactions after the rate-determining
step.
As an example, consider the following mechanism composed of unimolecular (elementary) reaction.
A
k1
k
1
When steps 2 B
B
K2
k
2
C
K3
k
D
3
(C) is assumed to be the rate-determining step. For this assumption to be valid,
we must have k –1 >> k2. The slow of B
C compared with B
A ensures that most B molecules go
back to A rather than going to C, thereby ensuring that step 1(A
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CHEMICAL KINETICS
Further-more, we must have k3 >> k2 and k3 >> k–2 to ensure that step 2 acts as a"bottleneck" and that
product D is rapidly formed from C. The overall rate is then controlled by the rate-determining step
B
C . (Note that since k3 >> k –2, the rate-limiting step is not in equilibrium.) Since we are examining
the rate of the forward reaction A
D , we further assume that k2[B] >> k–2[C]. During the early stage
of the reaction, the concentration of C will be low compared with B, and this condition will hold. Thus
we neglect the reverse reaction for step 2. Since the rate-controlling step is taken to be essentially
irreversible, it is irrelevant whether the rapid steps after the rate-limiting step are reversible or not.
The observed rate law will, depends only on the nature of the equilibria that precede the ratedetermining step and on this step itself.
ARRHENIUS EQUATION
Number of effective collisions = Number of collision × fraction.
k = A e–Ea/RT
Where ,
k
A
= rate constant of reaction
= Pre exponential factor or Arrhenius constant or Frequency factor.
Ea = Activation energy
R
= Universal gas constant
T
= Absolute temperature.
Fraction of molecules undergoing effective collision = e–Ea / RT =
K
A
Variation in Arrhenius equation
Type -1 :
log k = 2 –
k
5 10 3
T
log k = log A –
Ea
2.303 RT
A = ?
Ea = ?
= Ae–Ea/RT
loge k = loge A –
log10 A = 2
;
Ea
RT
Ea
2.303 RT
A = 102 = 100
5 10 –3
T
Ea = 2.303 × R × 5 × 10–3
Type -2 : Temperature Variation :
r1
K1
T1
r2
K2
T2
r2
r1
k2
k1
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Ea
Ea
log k1 = log A – 2.303 RT, log k2 = log A – 2.303 RT
1
2
log
k2
k1
Ea
1
1
–
2.303 R T1
T2
Type -3 :
Addition of Catalyst :
Activation Energy
Without Catalyst
Activation Energy
With Catalyst
Reaction Path
Without Catalyst
Reaction Path
Without Catalyst
Reaction
H
Products
REACTION COORDINATE
r1
k1
r2
log k 2 = log A –
k2
k1
Ea
T
catalyst
Eac
T
k1
log k1 = log A –
log
no catalyst
Ea
2. 303 RT
Eac
2. 303 RT
Ea – Eac
2.303 RT
Type -4 : Both Catalyst and Temperature :
log
k2
k1
1
Ea Eac
–
2.303 R T1
T2
Problem : 8
At 278 °C the half life period for the first order thermal decomposition of ethylene oxide is 363
min and the energy of activation of the reaction is 52,00 cal/mole. From these data estimate
the time required for ethylene oxide to be 75% decomposed at 450°C.
k 450
5200 1
1
–
1122
.
Sol.
ln k
= 2
551 725
278
k 450
k 278
363
3.07 =
t 1/ 2 (at 450 C)
Now t 0.75
1
A0
ln
k A0 / 4
1
ln 4
k
t1/2 (at 450°C) = 118.24 min.
1386
.
k
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t0.75 =
1386
.
× 118.24 = 236.48 min
0.693
Problem : 9
The act ivat ion energy of t he react ion : A + B
Sol.
products is 102.9 kJ/mol. At 40°C, the products
are formed at the rate of 0.133 mol/L/min. What will be rate of formation of products at 80°C?
Let the rate law be defined as
At T1 : r1 = k 1[A]x[B]y
At T2 : r2 = k2[A]x [B]y
r2 = r 1
k2
k1
Using Arrhenius equation find k at 40°C.
k2
Ea
T2 – T1
log10 k = 2.303R T T
1
1 2
log10
102.9 10 3
40
2.303 8.31 313 353
k2
k1
k2
log10 k1 = 1.95
k2
k1
88.41
r2 = 0.133 × 88.41 = 11.76 mol/L/min
Problem : 10
The activation energy of a non-catalysed reaction at 37°C is 200 kcal/mol and the activation
energy of the same reaction when catalysed decreases to only 6.0 kcal/mol. Calculate the
ratio of rate constants of the two reactions.
Sol.
We know that :
k
A e – E a / RT
Let k = rate constant for non-catalysed reaction and kc rate constant for catalysed reaction. Let Ea be
the activation energy for non-catalysed reaction and Eac be the energy of activation of catalysed
reaction.
1.
k
2.
kc
A e–E
k
kc
eRT
log10
k
kc
1
( 6 10 3 – 200 103 )
2.303 2 310
log10
k
kc
–9.8
A e – E a / RT
1
/ RT
(E –Ea )
log10
k
kc
k
kC
1
(Eac – Ea )
2.303 RT
156
.
10 –10
or
kc
k
6.3 10 9
Problem : 11
A first order reaction A B requires activation energy of 70 kJ/mol. When 20% solution of A
was kept at 25°C for 20 minutes, 25% decomposition took place. What will be the percent
decomposition in the same time in a 30% solution maintained at 40°C? Assume that the
activation energy remains constant in this range of temperature.
Sol.
Note : It does not matter whether you take 20%, 30%, 40% or 70% of A.
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At 25°C, 20% of A decomposes 25%
k t = 2.303 log10
A0
A
100
k(at 25° C) = 0.0143 min–1
75
Using Arrhenius equation find k at 40°C.
k(40) = 2.303 log10
k40
k25
log10
log10
Ea
T2 – T1
2. 303R T1T2
70 103
40 – 25
2.303 8.31 298 313
k 40
0.0143
k(at 40°C) = 0.055 min–1
Now calculate % decomposition at 40°C using first order kinetics.
kt = 2.303 log10
100
100 – x
0.055 × 40 = 2.303 log10
x = 67.1
A0
A
67.1% decomposition of A at 40°C.
Problem : 12
The rate constant of a reaction is 1.5 × 107 sec–1 at 50°C and 4.5 × 107 sec–1 at 100°C. Evaluate
the Arrhenius parameters A and Ea.
Sol.
2.303 log10
2.303 log10
Ea T2 – T1
K2
= R
T1T2
K1
4.5 107
15
. 107
=
373 – 323
Ea
8.314 373 323
Ea = 2.2 × 104 J mol–1
4
Now,
K
Ae –Ea /RT
4.5 × 107 =
Ae
–
2. 2 10
8.314 373
A = 5.42 × 1010
Problem : 13
Sol.
A reaction proceeds five times more at 60°C as it does at 30°C. Estimate energy of activation.
Given, T2 = 60 + 273 = 333 K, T1 = 30 + 273 = 303 K,
R = 1.987 × 10–3 kcal
r = k [ ]n
(at a temperature T)
r2
r1
(at temp. T2 and T1)
K2
K1
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CHEMICAL KINETICS
r2
r1
K2
K1
5
K2
2.303 log10 K
1
5
Ea [ T2 – T1 ]
R T1T2
Ea
333 – 303
333 303
10 –3 1987
.
Ea = 10.757 kcal mol–1
2.303 log10 5 =
Problem : 14
The rate constant of a reaction increases by 7% when its temperature is raised from 300 K to
310 K, while its equilibrium constant increases by 3%. Calculate the activation energy of the
forward and reverse reactions.
Sol.
Rate constant at 300K = k
7
100 = 1.07 k
Rate constant at 310 K = k + k
Thus,
2.303 log
k2
k1
2.303 log
107
. k
k
Eaf [ T2 – T1]
R T2 T1
Eaf 310 – 300
2 310 300
Eaf = 1258.68 cal
Now, equilibrium constant at 300 K = K
Equilibrium constant at 310 K = K +
3
× K = 1.03 K
100
K2
H T2 – T1
2.303 log K = R
T1T2
1
Using
2.303 log
103
. K'
K'
H 310 – 300
2 310 300
H = 549.89 cal
Since,
H = Eaf – Eba
549.89 = 1258.68 – Eba
Eba = 708.79 cal
Problem : 15
At 380°C, the half life period for the first order decomposition of H2O2 is 360 min. The energy of
activation of the reaction is 200 kJ mol–1. Calculate the time required for 75% decomposition at
450°C.
[IIT 1995]
Sol.
K1 = 0.693/360 min–1 at 653 K and
Ea = 200 × 103 J,
K 2 = ? at 723 K,
R = 8.314 J
From 2.303 log10 (K2 / K1) = (Ea /R)[(T2 – T1)/(T1T2)]
K2 = 0.068 min–1
Now,
t=
2.303
100
log10
= 20.39 minute
0.068
25
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OBJECTIVE SOLVED PROBLEMS
Problem : 1
In gaseous reactions important for understanding of the upper atmosphere H2O and O react
bimolecularly to form two OH radicals. H for this reaction is 72kJ/mol at 500 K and Ea = 77 kJ/
mol, then Ea for two bimolecular recombination of 2OH radicals to form H2O & O is
(A) 3 kJ mol–1
(B) 4 kJ mol–1
–1
(C) 5 kJ mol
(D) 7 kJ mol–1
Sol.
As H is positive, therefore reaction is endothermic
This is the energy profile diagram for an endothermic reaction.
Now when the products is converted back to reactant the
energy of activation is x as shown in fig.
Evidently x = Ea – H
= (77 – 72) = 5 kJ mol–1
(C)
E
x
Ea
H
Reaction co-ordinate
Problem : 2
In a certain reaction 10% of the reactant decomposes in the first hour, 20% is second hour,
30% in third hour and so on. What are the dimensions of rate constant.
(A) hour–1
(B) mol lit–1 sec–1
(C) lit mol–1 sec –1
(D) mol sec–1
Sol.
If the amount of products formed which is 10%, 20% and 30% is plotted against time i.e., 1 hr, 2 hr and
3 hr respectively, it is a straight line passing through the origin.
it is a zero order reaction where x = kt
x
=k
t
dimensions of k = moles lit–1 sec–1
(B)
Problem : 3
Two substances A(t½ = 5 mins) and B(t½ = 15 mins) are taken is such a way that initially [A]
= 4[B]. The time after which the conentration of both will be equal is
(A) 5 min
(B) 15 min
(C) 20 min
(D) concentration can never be equal
Sol.
t½ of A is 5 min
in 15 mins it will become 1/8 of initial and t½ of B is 15 mins
in 15 mins it will become ½ of initial
ratio of [A] : [B] after 15 min is 4 : 1
But given [A] = 4[B]
[A] = [B] after 15 min
[B]
Problem : 4
The reaction A(g) + 2B(g) C(g) + D(g) is an elementary process. In an experiment, the initial
partial pressure of A & B are PA = 0.60 and PB = 0.80 atm. When Pc = 0.2 atm the rate of reaction
relative to the initial rate is
(A) 1/48
(B) 1/24
(C) 9/16
(D) 1/6
Sol.
A(g) +
2B(g)
C(g) +
D(g)
t = 0 0.60
0.80
0
0
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CHEMICAL KINETICS
t=t
(0.6 – 0.2)
(0.8 – 2 × 0.2)
(Rate)i = k[A] [B] = k[0.6][0.8]
2
0.2
0.2
2
(Rate)t = k[0.4] [0.4]2
k[0.4] 3
Rt
Ri
k[ 0.6][ 0.8]2
=
1
6
[D]
Problem : 5
For a hypothetical reaction A + B
C + D, the rate = k[A]–1/2[B]3/2. On doubling the concentration
of A and B the rate will be
(A) 4 times
Sol.
k = k[2]
–1/2
(B) 2 times
[2]
3/2
= k[2]
3/2–1/2
(C) 3 times
(D) none of these
= k = [2] = 2k
1
[B]
Problem : 6
An organic compound A decomposes by following two parallel first order mechanisms :
k1
B
k1 1
A
; k = 2 , k1 = 0.693 hr –1
k2
2
C
Select the correct statement(s)
(A) If three moles of A are completely decomposed then 2 moles of B and 1 mole of C will be
formed.
(B) If three moles of A are completely decomposed then 1 moles of B and 2 mole of C will be
formed.
(C) half life for the decomposition of A is 20 min
(D) half life for the decomposition of B is 0.33 min
Sol.
BC
k
k1
0.693
t1/2
k2
0.693 2 0.693
3 0.693
t1/2
1
hr
3
20 min
Problem : 7
For any Ist order gaseous reaction A
2B Pressure devoloped after 20 min is 16.4 atm and
after very long time is 20 atm. What is the total pressure devoloped after 10 min.
(A) 12
Sol.
(B) 13
(C) 14
(D) 15
C
A(g)
t=0
t = 10 PO - P
2B(g)
PO
2P
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t = 20 PO - P'
Page # 89
2P'
–
2PO = 20
2PO
PO = 10
16.4 = PO - P' + 2P'
16.4 = 10 + P'
P' = 6.4
for first order reaction at equal time conc/pr is in G. P.
Po
Po
P'
p
Po P 10 6 .4
,
Po
10 P
10 P
10
3.6 × 10 = (10 – P)2
10 – P = 6
P=4
P10 min = PO + P = 10 + 4 = 14 atm
Problem : 8
For any reaction, 2A
B, rate constant of reaction is 0.231 min–1. Time (in sec) when 25% of
A will remain unreacted.
(A) 150
Sol.
(B) 180
(C) 200
(D) 140
B
kr
kA
,
2
t75%
2t1 /2
0.693
2kr
2
= 3 mm
= 180 sec
Problem : 9
For any reaction A(g)
B(g), rate constant k = 8.21 × 10–2 atm/min at 300 K. If initial concen-
tration of A is 2M then what is the half life (in hr.)?
(A) 5
Sol.
(B) 6
(C) 7
(D) 8
A
Rate = k =
t1/2 =
a
2k
8.21 10 2 atm
0.0821 300 min
2
2k
1 mole
300 . min
1
= 300 min
k
= 5 hr.
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CHEMICAL KINETICS
Problem : 10
2A
B+C
The mechanism of the above reaction given as
K1
2A
X +B
Sol.
X + 2B (fast)
K2
K3
C (slow)
E1 = Activation energy for K1
E2 = Activation energy for K2
E3 = Activation energy for K3
Calculate Ereaction
(given : E1 = 10 kJ, E3 = 5 kJ, E2 = 12 kJ)
(A) 5
(B) 6
(C) 7
D
rate = K3 [x][B]
k1
k2
[X][B]2
[A]2
rate = k3 .
Keff =
(D) 3
k1 [A]2
, [X] = k
2
2 [B]
k 1 [A]2 [B]
k1 [A]2
.
,
=
k
.
.
3
k 2 [B]2
k 2 [B]
k 3 .k 1
k 2 , Etotal = E1 + E3 – E2
= 3kJ
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SUBJECTIVE SOLVED PROBLEMS
Problem : 1
The rate of change of concentration of C in the reaction 2A + B 2C + 3D was reported as 1.0 mol
litre–1 sec–1. Calculate the reaction rate as well as rate of change of concentration of A, B and D.
Sol.
We have,
–
1 d[ A]
d[B]
1 d[ C]
1 d[D]
= –
=
=
= rate of reaction
2 dt
dt
2 dt
3 dt
d[ C]
= 1.0 mol litre–1 sec–1
dt
–
d[ A]
d[ C]
=
= 1.0 mol L–1sec–1
dt
dt
–
d[B]
1 d[ C]
1
=
=
= 0.5 mol L–1 sec–1
dt
2 dt
2
d[D ]
dt
3 d[C]
3
1 = 1.5 mol L–1 sec –1
=
2 dt
2
Also,
Rate =
1 d[ C]
2 dt
Rate =
1
× 1 = 0.5 mol L–1 sec–1
2
Problem : 2
For the reaction A + B C, the following data were obtained. In the first experiment, when the
initial concentrations of both A and B are 0.1 M, the observed initial rate of formation of C is 1 ×
10–4 mol litre–1 minute–1. In the second experiment when the initial concentrations of A and B
are 0.1 M and 0.3 M, the initial rate is 9.0 × 10–4 mol litre–1 minute–1.
(a) Write rate law for this reaction
(b) Calculate the value of specific rate constant for this reaction.
Sol.
Let
Rate = K[A]m[B]n
(a)
r1 = 1 × 10–4 = K[0.1]m [0.1]m
...(1)
r2 = 9 × 10–4 = K[0.1]m[0.3]n
...(2)
r3 = 2.7 × 10–3 = K[0.3]m[0.3]m
...(3)
By Eqs. (1) and (2),
r1 1 10 –4
r2 9 10 –4
n= 2
By Eqs. (2) and (3),
1
3
n
m
(b)
r2
9 10 –4
1
m=1
r3 27 10 –4
3
Rate = K[A]1[B]2
Also by Eq. (1),
1 × 10 –4 = K[0.1]1 [0.1]2
K = 10–1 = 0.1 L2 mol–2 min–1
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CHEMICAL KINETICS
Problem : 3
The chemical reaction between K2C2O4 and HgCl2 is ;
2HgCl2 + K2C2O4 2KCl + 2CO2 + Hg 2Cl2
The weights of Hg2Cl2 precipitated from different solutions in given time were taken and
expressed as following :
Time (minutes)
HgCl2 (M)
K 2C2 O4 (M)
Hg 2Cl 2 formed (M)
60
0.0418
0.404
0.0032
65
0.0836
0.404
0.0068
120
0.0836
0.202
0.0031
Let the rate law be written as : r = k[HgCl2 ]x [K2C 2O4]y
1.
0.0032
= k[0.0418]x [0.404]y
60
2.
0.0068
= k[0.0836]x[0.404]y
65
0.0031
= k[0.0836]x [0.202]y
120
Solving the above equations, we get :
x = 1 and y = 2
order of reaction w.r.t x = 1 and y = 2 and overall order is 3.
Problem : 4
The reaction given below, involving the gases is observed to be first order with rate constant
7.48 × 10–3 sec–1. Calculate the time required for the total pressure in a system containing A at
an initial pressure of 0.1 atm to rise to 0.145 atm also find the total pressure after 100 sec.
2A(g)
4B(g) + C(g)
Sol.
2A(g)
4B(g) +
C(g)
initial
P0
0
0
at time t
P0 – P
2P
P /2
3.
Ptotal = P0 – P + 2P + P /2 = P0 +
P =
k=
2
(0.145 – 0.1) = 0.03 atm
3
2.303
P
log 0
2t
P0 – P'
2. 303
t=
7.48 10 – 3 2
t = 23.84 sec
Also,
k=
3P'
2
log
0. 1
0.07
2.303
0. 1
log
2t
P0 – P'
7.48 × 10–3 =
2.303
0.1
log
2 100
0.1– P'
0.1/0.1 – P = 5
P = 0.08
3
Ptotal = 0.1 + (0.080)
2
~
– 0.22 atm.
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CHEMICAL KINETICS
Page # 93
Problem : 5
The net rate of reaction of the change :
[Cu(NH3)4]2+ + H 2O
dx
dt
[Cu(NH3)3H2O]2+ + NH 3 is,
2.0 10 – 4 [Cu(NH ) ]2+ – 3.0 × 105 [Cu(NH ) H O]2+ [NH ]
3 4
3 3 2
3
calculate :
(i) rate expression for forward and backward reactions.
(ii) the ratio of rate constant for forward and backward reactions.
(iii) the direction of reaction in which the above reaction will be more predominant.
Sol.
(i) Rate of forward reaction = 2.0 × 10–4 [Cu(NH3)4]2+ [H2O]
Rate of backward reaction = 3.0 × 105 [Cu(NH3)3H2O]2+ [NH3]
(ii) Also,
K f = 2.0 × 10–4
Kb = 3.0 × 105
Kf
2.0 10 –4
Kb
3.0 10 5
= 6.6 × 10–10
(iii) More predominant reaction is backward reaction.
Problem : 6
The rate law for the decomposition of gaseous N2O5,
N2O5(g)
2NO2(g) +
1
O (g)
2 2
is observed to be
d[N2 O5 ]
= k[N2O5]
dt
A reaction mechanism which has been suggested to be consistent with this rate law is
r= –
N2O5(g)
k
NO2(g) + NO3 (g)
(fast equilibrium)
NO2 ( g) NO3 ( g)
k1
NO2(g) + NO(g) + O2(g)
(slow)
NO(g) + NO3(g)
k2
2NO2(g)
(fast)
Show that the mechanism is consistent with the observed rate law.
Since the slow step is the rate determining step, hence
r = k1[NO2] [NO3]
...(1)
and from the fast equilibrium step,
K
[NO2 ][NO 3 ]
[N2 O 5 ]
Thus, [NO2] [NO3] = K[N 2O5]
...(ii)
Using (ii) in (i), we get :
r = k1K[N2O5] = k[N2 O5]
where k = k1K
This shows that the mechanism is consistent with the observed rate law.
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CHEMICAL KINETICS
Problem : 7
The half life of first order decomposition of nitramide is 2.1 hour at 15°C.
NH2NO2(aq)
N2O(g) + H2O (l)
If 6.2 gm of NH2NO2 is allowed to decompose, find :
Sol.
(a)
time taken for nitramide to decompose 99%;
(b)
volume of dry N2O gas produced at this point at STP.
(a)
Using first order kinetics, we have :
kt = 2.303 log10
0.693
21
.
(b)
A0
A
t = 2.303 log
100
100 – 99
6.2 gm of NH2NO2
and
1 mole NH2NO2
t = 13.96 hours
0.1 mol
1 mole of N2O
As 99% of NH2NO2 is decomposed
0.099 mol of NH2NO2 is decomposed
0.099 mol of N2O are produced
22.4 × 0.099 = 2.217 L of N2O at STP.
Problem : 8
The reaction A + OH–
–d[A]
dt
Products, obeys rate law expression as,
k[A][OH– ]
If initial concentrations of [A] and [OH –] are 0.002 M and 0.3 M respectively and if it takes 30
sec for 1% A to react at 25°C, calculate the rate constant for the reactions.
Sol.
A
t=0
+
OH–
0.002
0.002 –
t = 30
Using K
0.3
0.002 1
100
0.3 –
0.002 1
100
2.303
b(a – x)
log10
t( a – b)
a(b – x)
0.002 1
100
0.002 1
0.002 0.3 –
100
0.3
K
Products
2.303
log10
30 ( 0.002 – 0.3)
0.002 –
K = 1.12 × 10–3 L mol–1 sec–1
Problem : 9
A certain reaction A + B products ; is first order w.r.t. each reactant with k = 5.0 × 10–3 M–1s–1.
Calculate the concentration of A remaining after 100s if the initial concentration of A was 0.1 M
and that of B was 6.0 M. State any approximation made in obtaining your result.
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Sol.
A+B
Page # 95
products
Given : Rate = k[A][B]
(2nd Order reaction)
Now, since [B] >> [A], [B] can be assumed to remain constant throughout the reaction. Thus, the rate
law for the reaction, becomes :
Rate
k0[A]
where k0 = k[B] = 5.0 × 10–3 × 6.0 s–1 = 3.0 × 10–2 s–1
Thus, the reaction is now of first order.
Using, 2.303 log10
2.303 log10
loge
e3
k 0t
A
0.1
k 0t
A
0.1
3
A
0.1
A
A0
3
[
logex = 2.303 log10x]
5 10 – 3 M
Problem : 10
Dimethyl ether decomposes according to the following reaction :
CH3 – O – CH3(g)
CH4(g) + CO(g) + H2(g)
At a certain temperature, when ether was heated in a closed vessel, the increase in pressure
with time was noted down.
Time (min)
0
10
20
30
Pressure (mm Hg)
420
522
602
678
(i)
Show that the reaction is first order.
(ii)
Compute the pressure of CO(g) after 25 minutes.
Sol.
CH3 – O – CH3 (g)
CH4(g) + CO(g) + H2 (g)
CH 3–O–CH3
time
C0
t=0
t=t
Ct
(all are gases)
CH4
CO
H2
P0
0
0
0
P0 – x
x
x
x
Pt = P0 + 2x
x=
A0
A
1
(P – P0)
2 t
P0
P0 – x
2P0
3P0 – Pt
Now find k1, k2 and k3 using the first order kinetics
2P0
k t = 2.303 log10 3P – 2P
0
t
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CHEMICAL KINETICS
k1 =
2( 420)
2.303
log10
= 0.0129 min–1
3
(
420
) – 522
10
k2 =
2( 420)
2.303
log10
= 0.0122 min–1
3
(
420
) – 602
20
k3 =
2( 420)
2.303
log10
= 0.0123 min–1
3( 420) – 678
30
As k1 ~ k2 ~k3 , the reaction is first order.
kaverage =
1
(k k 2
3 1
PCO = x =
k 3 ) = 0.0127 min–1
1
(Pt – P0 )
2
Find P after t = 25 min using first order kinetics with k = 0.0127 min–1
log10
2( 420)
3(240) – Pt
Pt = 648.46 mm
0.0127 25
x = 114.23 mm
Problem : 11
The decomposition of N2O5 according to following reaction is first order reaction :
2N2O5(g)
4NO 2(g) + O2 (g)
After 30 min. from start of the decomposition in a closed vessel, the total pressure developed
is found to be 284.5 mm of Hg and on complete decomposition, the total pressure is 584.5 mm
of Hg. Calculate the rate constant of the reaction.
Sol.
2N2O5(g)
4NO2(g) + O2(g)
2N2O5
4NO2
O2
P0
0
0
P0 – 2x
4x
x
P0 : initial pressure ; Let Pt : pressure at 30 min and P : pressure at the end of decomposition.
Pt = P0 + 3x
and
P = 2P0 +
x=
1
5
P0 = P0
2
2
1
(P – P0)
3 t
P0 =
2
P
5
For the first order kinetics
keff t = 2.303 log10
A0
A
A0 : initial concentration ;
A : final concentration
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CHEMICAL KINETICS
A0
A
Now
keff =
Page # 97
1
P
5
1
P – 2 / 5P
P –2 t
5
3
P0
P0 – 2 x
A0
A
3
P
5 P – Pt
1
3
584.5
× 2.303 log10
×
= 5.204 × 10–3 min–1
30
5
584.5 – 284.5
k for the reaction =
5.204
2
10 –3 = 2.602 × 10–3 min–1
Problem : 12
The gas phase decomposition of N2O5 to NO2 and O2 is monitored by measurement of total
pressure. The following data are obtained.
P total (atm)
0.154
0.215
0.260
0.315
0.346
Time (sec)
1
52
103
205
309
Find the average rate of disappearance of N2O5 for the time interval between each interval and
for the total time interval. [Hint : Integrated rate law is NOT to be used]
Sol.
2N2O5(g)
Initial Pressure (at t = 0)
At equilibrium
Now:
4NO2(g) + O2(g)
P0
P0 – 2x
0
4x
Pt = (P0 – 2x) + 4x + x
PN 2O 5
PN2 O 5
Thus,
P0 – 2x
0
x
x=
1
(P – P0 )
3 t
1
( 5P0 – 2Pt )
3
2
(Pt 1 – Pt 2 ) where Pt 2 and Pt 1 are the total pressures at time instants t2 and t1 (t2 > t1)
3
respectively
Ptotal (atm) Time (sec)
0.154
1
0.215
52
0.260
103
0.315
205
0.346
309
PN2 O5
t
= Avg. Rate of disappearance of N2 O5
( 0.154 – 0.215)
(52 – 1)
( 0.215 – 0.260)
(103 – 52)
( 0.260 – 0.315)
(205 – 103)
( 0.315 – 0.346)
( 309 – 205)
–1.20 10 –3
–0.88 10 –3
–0.54 10 –3
–0.30 10 –3
Problem : 13
5 ml of ethylacetate was added to a flask containing 100 ml f 0.1 N HCl placed in a thermostat
maintained at 30°C. 5 ml of the reaction mixture was withdrawn at different intervals of time
and after chilling, titrated against a standard alkali. The following data were obtained :
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CHEMICAL KINETICS
Time (m inutes)
0
75
119
183
Volum e of alkali used in m l
9.62
12.10
1.10
14.75
21.05
Show that hydrolysis of ethyl acetate is a first order reaction.
Sol.
The hydrolysis of ethyl acetate will be a first order reaction if the above data confirm to the equation.
k1 =
V – V0
2.303
log
t
V – Vt
Where V0, Vt and V represent the volumes of alkali used at the commencement of the reaction, after
time t and at the end of the reaction respectively, Hence
V – V0 = 21.05 – 9.62 = 11.43
Time
V – Vt
k1
75 min
21.05 – 12.10 = 8.95
2.303
1143
.
log
75
8.95
= 0.003259 min–1
119 min
21.05 – 13.10 = 7.95
2.303
1143
.
log
= 0.003051 min–1
119
7.95
183 min
21.05 – 14.75 = 6.30
2.303
1143
.
log
= 0.003254 min–1
183
6.30
A constant value of k shows that hydrolysis of ethyl acetate is a first order reaction.
Problem : 14
The optical rotations of sucrose in 0.5N HCl at 35°C at various time intervals are given below.
Show that the reaction is of first order :
Sol.
Tim e (m inutes)
0
10
20
30
40
Rotation (degrees)
+32.4
+28.8
+25.5
+22. 4
+19. 6
–11.1
The inversion of sucrose will be first order reaction if the above data confirm to the equation, k1 =
r0 – r
2.303
log r – r
t
t
Where r0, rt and r represent optical rotations initially, at the commencement of the reaction after time
t and at the completion of the reaction respectively
In the case a0 = r0 – r = +32.4 – (–11.1) = +43.5
The value of k at different times is calculated as follows :
Time
rt
rt – r
k
2.303
43.5
log
10 min
+28.8
39.9
= 0.008625 min–1
10
39.9
2.303
43.5
log
20 min
+25.5
36.6
= 0.008625 min–1
10
36.6
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30 min
Page # 99
+22.4
2.303
43.5
log
= 0.008694 min–1
30
33.5
33.5
2.303
43.5
log
= 0.008717 min–1
40
30.7
The constancy of k1 indicates that the inversion of sucrose is a first order reaction.
Problem : 15
The hydrolysis of ethyl acetate
40 min
+19.6
30.7
CH3COOC2H5 + H2O
Sol.
CH3COOH + C2H5OH
in aqueous solution is first order with respect to ethyl acetate. Upon varying the pH of the
solution the first order rate constant varies as follows.
pH
3
2
1
k1 ×10–4s–1 1.1
11
110
what is the order of the reaction with respect of H+ and the value of the rate constant?
Rate = k[CH3COOC2H5]a[H+]b
[H+] is constant through out the reaction
k1 = k[H+]b
Hence,
k1
[H ]1
k 1'
[H ] 2
b= 1
k1= k[H+]
1.1 × 10–4 = k(10–3 )
b
11
.
11
10 –3
b
10 –2
k = 1.1 × 10–1 dm3 mol–1 sec–1
Problem : 16
Two I order reactions having same reactant concentrations proceed at 25°C at the same rate.
The temperature coefficient of the rate of the first reaction is 2 and that of second reaction is 3.
Find the ratio of the rates of these reactions at 75°C
Sol.
For I order reaction r1 = K[C]1
R1
R2
K1 / K 2 = temperature coefficient
Let temperature co-efficient be a
R35
R25
K 35
K 25
a
R45
R25
a a
a2
R75
Similarly, R
25
R 45
R 35
K 45
K 35
a
a5
For I reaction (R75)I = 25 × (R 25)I
For II reaction (R75)II = 35 × (R 25)II
(R75 )II
(R75 )I
35
25
= 7.9537
[
(R25)I = (R25)II]
Problem : 17
For the reaction :
C2H5I + OH– C2H5OH + I–
the rate constant was found to have a value of 5.03 × 10–2 mol–1 dm3 s–1 at 289 K and 6.71 mol–
1
dm3 s–1 at 333 K. What is the rate constant at 305 K.
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Sol.
CHEMICAL KINETICS
k2 = 5.03 × 10–2 mol–1 dm3 s–1 at T2 = 289 K
k1 = 6.71 mol–1 dm3 s–1 at T1 = 333 K
6.71
log
–2
Ea
333 – 289
= 2.303 8.314 333 289
5.03 10
On solving we get, Ea = 88.914 kJ
The rate constant at 305 K may be determined from the relation :
log
k1
Ea
1
1
–
k 2 = 2.303R T2 T1
log
k1
5.03 10 –2
88.914
1
1
= 2.303 8.314 298 – 305
On solving we get, k1 = 0.35 mol–1 dm3 s –1
Problem : 18
A secondary alkyl halide (RX) is hydrolysed by alkali simultaneously by SN1 and SN2 pathways. A plot of
1 d RX
vs [OH –] is a straight line with slope equal to 1.0 × 10–3 mol –1 L min –1 and intercept equal to
RX dt
Sol.
2.0 × 1 0–3 min –1. Calculate initial rate of consumption of RX when [RX] = 0.5 M and
[OH–] = 1.0 M.
1.5 × 10–3 mol L–1 min–1
For SN1 pathway:
d RX
K1[RX] K1 = 1st order rate constant
dt
For SN2 pathway:
d RX
= K2[RX][OH–]
K 2 = 2nd order rate constant
dt
Thus, the overall rate of consumption of RX is as given below:
d RX
= K1[RX] + K2[RX][OH–]
dt
or –
1 d[RX]
= K 1 + K2 [OH–]
[RX] dt
According to this equation as plot of –
1 d[RX]
vs [OH|–] will be a straight line of the slope equal
[RX] dt
to K2 and intercept equal to K1 . Thus, from question.
K1 = 2.0 ×10–3 min–1
K2 = 1.0 ×10–3 min–1 L min–1
d[RX]
= 2.0 × 10–3 × 0.5 + 1.0 × 10–3 × 0.5 × 1
dt
= 1 × 10–3 + 0.5 × 10–3
= 1.5 × 10–3 mol L–1 min–1
Thus, –
Problem : 19
A polymerisation reaction is carried out at 2000 K and the same reaction is carried out at 4000 K with
catalyst. The catalyst increases the potential barrier by 20 KJ but the rate of the reaction remains
same. Find activation energy of the reaction.
(Assuming all other parameters to be same.)
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Sol.
Page # 101
0020
k = Ae–Ea/R×2000 =
Ea
R 2000
Ae
Ea 20
R 4000
(Ea 20)
R 4000
2Ea = Ea + 20
Ea = 20 kJ
Problem : 20
3k
3B
Consider the following first order parallel reaction. 2A
5k
5C
The concentration of C after time t is :
Sol.
25
[1 – e –16kt ]
16
–
1 dA
2 dt
–
dA
dt
1 dB
3 dt
8kA
1 dC
= 3k[A] + 5k[A] = 8kA
5 dt
2 = 16 kA
A = A0 e–16 kt
1 dB
3 dt
3kA ,
B=
dB
= 9kA = 9kA0 e–16kt
dt
9
[1 – e–16kt ]
16
Similarly
25
[1 – e–16kt ]
16
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CHEMICAL KINETICS
Class Room Problems
Ex.1 The oxidation of iodide ion by peroxy disulphate
ion is described by the equation :
3I– + S2O82–
(a) If –
2–
2 8
I3– + 2SO42–
= 1.5 × 10–3 Ms–1 for a particular
time interval, what is the value of –
[I– ]
for the same
t
time interval ?
(b) What is the average rate of formation of SO42–
during that time interval ?
Sol.
(a) 4.5 × 10–3 (b) 3 × 10–3
Ex.2 In the following reaction 2H2O2
rate of formation of O2 is 36 g min–1,
Ex.3 The stoichiometric equation for the oxidation of
bromide ions by hydrogen peroxide in acid solution is
2Br– + H2O2 + 2H+ Br2 + 2H2O
Since the reaction does not occur in one stage, the
ra te eq uatio n d oes not c orresp ond to thi s
stoichiometric equation but is
rate = k[H2O2] [H+] [Br–]
(a) If the concentration of H2O2 is increased by a
factor of 3, by what factor is the rate of consumption
of Br– ions increased ?
(b) If, u nder certai n c on diti ons , the rate of
consumption of Br– ions is 7.2 × 10–3 mole dm–3 s–1,
what is the rate of consumption of hydrogen peroxide
? What is the rate of production of bromine
(c) What is the effect on the rate constant k of
increasing the concentration of bromide ions ?
(d) If by the addition of water to the reaction mixture
the total volume were doubled, what would be the
effect on the rate of change on change on the
concentration of Br– ? What would be the effect on
the rate constant k ?
Sol. (a) 3 times (b) 3.6 × 10–3 (c) No. effect
(d) 1/8 times, no. effect on K
2H2O + O2
(A) What is rate of formation of H2O ?
(B) What is rate of disappearance of H2O2 ?
Sol.
(A) 40.5 (B) 76.5
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Ex.4 For a reaction 2P + Q S ; following data were
collected. Calculate the overall order of the reaction.
Also find out the reaction rate constant.
P(mole/L)
Q(mole/L) Rate
–3
1. 6 × 10
1 × 10–3
0.012
–3
2. 6 × 10
2 × 10–3
0.024
3. 2 × 10–3
1.5 × 10–3 0.0024
4. 4 × 10–3
1.5 × 10–3 0.0096
Sol.
m = 2, n = 1
Ex.5A 22.4 litre flask contains 0.76 mm of ozone at
25ºC. Calculate
(i) the concentration of oxygen atom needed so that
the reaction O + O3 2O2 having rate constant equal
to 1.5 × 107 litre mol–1 sec–1 can proceed with a rate
of 0.15 mol litre–1 sec–1.
(ii) the rate of formatiuon of oxygen under this condition.
Page # 103
Ex.6 The rate of a reaction A + B
studied to give following data.
products is
Initial [A] in mol/L
Initial [B] in mol/L
ra te (m ol/L/min)
0.01
0.01
0.005
0.02
0.01
0.010
0.01
0.02
0.005
What is the rate law ? What is the half-life of A in the
reaction
Sol. K = 0.5, t1/2 = 1.386 min, m = 1, n = 0
Ex.7 For the reaction :
2A + B + C
A2B + C
2
the rate = k[A][B] with k = 2.0 × 10–6 M–2 s –1.
Calculate the initial rate of the reaction when [A] =
0.1 M, [B] = 0.2 M and [C] = 0.8 M. If the rate of
reverse reaction is negligible then calculate the rate
of reaction after [A] is reduced to 0.06 M.
Sol. 2.8 × 10–9 , 3.88 × 10–9
Sol. (i) 2.4 × 10–4 (ii) 2 × 0.15 Ms–1 = 0.30
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CHEMICAL KINETICS
Ex.8 A & B are two different chemical species
undergoing 1st order decomposition with half lives equal
to 0.231 & 0.3465 sec respectively. If the initial conc.
[ A]
of A & B are in the ratio 3 : 2. Calculate
after
[B]
theree half lives of 'A'.
Sol.
0.375 : 0.5
Ex.10 Inversion of sucrose is studied by measuring
angle of rotation at any time t.
C12H22O11 (sucrose) + H2O
C 6H12O6 (glucose)
+ C6H12O6 (fructose)
It is found that are (r0 – r ) a and (rt – r ) (a
- x) where r0, rt and r
are the angle of rotation at
the start, at the time t and at the end of the respectively. From the following values calculate the rate
constant and time at which solution is optically inactive.
Time/ min.
0.0
46.0
Rotation of Glucose
& Fructose
Sol. t = 104.43 min
Ex.9A
24.1
10.0
–10.7
B+C
Time
T
Volume of reage nt
V2
V3
Reagent reacts with all A, B and C and have 'n' factors
in the ratio of 1 : 2 : 3 with the reagent. Find k.
4V3
Sol. K = 1/t ln 5V 5V
3
2
Ex.11 The first order reaction :sucrose
Glucose + Fructose takes place at 3080 K in 0.5 NHCl. At
time zero the initial total rotation of the mixture is
32.40. After 10 minutes the total rotation is 28.80. If
the rotation of sucrose per mole is 850, that of glucose of the glucose is 740 and fructose is -86.040.
Calculate half life of the reaction.
Sol.
0.3 min
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CHEMICAL KINETICS
Ex.12 At room temperature (200C) milk turns sour in
about 65 hours. In a refrigerator at 30C milk can be
stored three times as long before its sours. Estimate
: (a) Activation energy of the reaction that causes
the souring of milk, (b) How long should it take milk to
sour at 400C.
Sol. 43.47 kJ/mole, 20.5 hr.
Page # 105
Sol.
Ex.15 The effective frequency factor
(a) A'
Ex.13 When the catalyst is removed from a reaction
at 5000K the rate decreases upto 4 times and activation energy is increased by 20%. Find the ratio of
velocity constants at temperature 3200K and 3000K
in the absence of catalyst. Assume activation energy
is independent to temperature.
K2
Sol. Ea = 34.6, K = 2.378
1
A2 A1
A3 A5
2
(c) A’= A1 × A2 × A3 × A4
(b) A'
2A2
A3
(d) A'
A1
A2
A1
A5
A3
A5
Sol.
Ex.16 Which one is correct ?
1/5
(a) k' 2
COMPREHENSION
For a reaction, if effective rate constant k’ is given
A2 A1
A 3 A51/ 5
exp
– Ea 2
exp
– Ea 1
exp
– Ea 1
1/5
(b) k' 2
A2 A1
A 3 A51/ 5
Ea 3
1/ 5Ea 1 – 1/ 5Ea 5
RT
Ea 3
1/ 5Ea 2 – 1/ 5Ea 5
RT
1/ 5
2k 2 k1
where k 1 , k 2 , k 3 , k 5 are rate
k 3 k5
constants for different steps of the reaction
by k =
(c) k' 2
A2 A5 1/5
A 3 A11/ 5
Ea 2
Ea 3
Ea 5
RT
(d) none of these.
Ex.14 The effective activation energy is
(a) E'a
Ea 2 – Ea 3
1
1
Ea – Ea
5 1 5 5
(b) E'a
Ea1
Ea 2
Ea 3
(c) E'a
Ea 1
Ea 2 – Ea 3 – Ea 5
(d) E'a
Ea 2
Ea 3 – Ea 5 – Ea 1
Sol.
Ea 5
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1/ 5
1/ 5
Page # 106
CHEMICAL KINETICS
Ex.17 A certain organic compound A decomposes by
two parallel first order mechanism
B
k
1
A
If k1 : k2 = 1 : 9 and k1 = 1.3 × 10 s .
C
Calculate the concentration ratio of C to A, if an ex–5
Ex.20. The reaction
cis – Cr(en)2 (OH)+2
–1
k2
periment is started with only A and allowed to run for
one hour.
Sol. 0.537
k1
k2
trans – Cr(en)2 (OH)+2
is first order is both directions. At 25°C the equilibrium
constant is 0.16 and the rate constant k1 is 3.3 × 10–
s . In an experiment starting with the pure cis form,
4 –1
how long would it take for half the equilibrium amount
of the trans isomer to be formed ?
Sol.
4.83 min.
B
k1
Ex.18 For a parallel first order reaction
k2
A
C
k3
where k1 = x hr–1 and k1 : k2 : k3 = 1 : 10 : 100.
D
[B]
[C]
[D]
Calculate [A ] , [A ] and [A ] = ...................
t
t
t
A0
A
Sol. Keff = 111 k1
A
B
C D
A
= e 111k1t
Ex.19 An organic compound A de composes following
two parallel first order mechanisms :
k1
A
k2
Ex.21 Two reactions (i) A
products (ii) B
products, follow first order kinetics. The rate of the
reaction (i) is doubled when the temperature is raised
from 300 K to 310 K. The half life for this reaction at
310 K is 30 minutes. At the same temperature B
decomposes twice as fast as A. If the energy of
activation for the reaction (ii) is half that of reaction
(i), calculate the rate constant of the reaction (ii) at
300 K.
Sol. 0.0327 min–1
B
k1 1
; k = 10 and k1 = a hr –1.
2
C
Calculate the concentration ratio of C to A after one
hour from the start of the reaction, assuming only A
was present in beginning.
Sol.
10 11 x
(e
11
1)
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CHEMICAL KINETICS
Ex.22 For a reaction A2 + B2
2AB, evaluate the
energy of activation from the following data:
Sol. 6×104 J, 9.3 ×104 J
1 –1
(K ) log10 K
T
T (in K)
500
200
Page # 107
2×10–3 3.0
5×10–3 2.0
Sol. Ea = 6.382 kJ
Ex.25 A first order reaction A B requires activation
energy of 70 kJ mol–1 . When a 20% solution of A was
kept at 25ºC for 20 minute, 25% decomposition took
place. What will be the percent decomposition in the
same time in a 30% solution maintained at 40ºC? Assume that activation energy remains constant in this
range of temperature.
Sol. 67.2%
Ex.23 A reaction proceeds five times more at 60ºC as
it does at 30ºC. Estimate energy of activation.
Sol. 45 kJ
Ex.26 At 380ºC, the half life period for the first order
decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ mol–1. Calculate the
time required for 75% decomposition at 450ºC.
Sol. 20.4
Ex.24 For the reaction A
B, E for the reaction is
–33.0 kJ mol–1. Calculate :
(a) equilibrium constant KC for the reaction at 300 K
(b) If Ef and Eb are in the ratio 20 : 31, calculate Ef, Eb
at 300 K. Assuming pre-exponential factor same for
forward and backward reactions.
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Page # 108
CHEMICAL KINETICS
Ex.27 The energy of activation for a reaction is 100
kJ mol–1. Presence of a catalyst lowers the energy of
activation by 75%. What will be effect on rate of
reaction at 20ºC; other things being equal?
Sol. 2.35 × 1013 time = e30.79
2 I + H2
(c) I2
I + H2
2HI (slow)
2I
I H2
IH2 + I
2 H I (slow)
(d) Can the observed rate law expression rate =
k[H2][I2] distinguish among these mechanisms ?
(e) If it is known that ultraviolet light causes the
reaction of H2 and I2 to proceed at 200° C with the
same rate law expression, which of these mechanisms
becomes most improbable ? Are any of these mechanisms proved ?
Sol. No, Mechanism 'a' is wrong.
Ex.28 (a) The reaction A proceeds in parallel chanB
nels A
Although the A
C branch is thermody-
C
namically more favorable than the branch A B, the
product B may dominate in quantity over C. Why may
this be so ?
(b) In the above problem, suppose the half life val-
Ex.30 Hydrolysis of methyl acetate in aqueous
solution has been studied by titrating the liberated
acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.
t/min
0
30
60
90
C/mol L–1 0.8500 0.8004 0.7538 0.7096
What is the value of k in this quation if
Rate = k [CH3COOCH 3][H2 O]
Ans. 3.64 × 10–5 M–1 min–1
ues for the two branches are 60 minutes and 90 minutes, what is the overall half-life value?
Sol.
Ex.29 Deduce rate law expressions for the conversion of H2 and I2 to HI at 400°C corresponding to each
of the following mechanisms :
(a) H2 + I2
(b) I2
2HI (one step)
2I
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CHEMICAL KINETICS
Page # 109
OBJECTIVE PROBLEMS (JEE MAIN)
EXERCISE – I
1. The rate of a reaction is expressed in different ways
as follows :
1 d[ C]
2 dt
–
1 d[D]
=
3 dt
The reaction is :
(A) 4A + B 2C + 3D
(C) A + B C + D
Sol.
1 d[ A]
4 dt
–
k1
k2
and
X
A B
Y
C D
if 50% of the reaction of X was completed when 96%
of the reaction of Y was completed, the ratio of their
d[B]
dt
(B) B + 3D
(D) B + D
4. Consider the following first order competing reactions :
4A + 2C
A+C
rate constants
(A) 4.06
Sol.
2. The rate con stant for the forward reaction
A(g)
2B(g) is 1.5 × 10–3 s–1 at 100 K. If 10–5
moles of A and 100 moles of B are present in a 10 litre
vessel at equilibrium then rate constant for the backward
reaction at this temperature is
(A) 1.50 × 104 L mol –1 s–1 (B) 1.5 × 1011 L mol –1 s–1
(C) 1.5 × 1010 L mol–1 s–1
(D) 1.5 × 10–11
Sol.
k2
k1
is
(B) 0.215
(C) 1.1
(D) 4.65
5. Units of rate constant for first and zero order
reactions in terms of molarity (M) are respectively.
(A) sec–1 , M sec–1
(B) sec–1, M
–1
–1
(C) M sec , sec
(D) M, sec–1
Sol.
6. For the reaction A + B
C; starting with different
initial concentration of A and B, initial rate of reaction
were determined graphically in four experiments.
3. Reaction A + B
C + D follow's following rate law :
rate = k[ A] 1/ 2 [B]1/2 . Starting with initial conc. of 1 M
of A and B each, what is the time taken for concentration of A of become 0.25 M.
Given : k = 2.303 × 10–3 sec –1.
(A) 300 sec
(B) 600sec
(C) 900 sec
(D) none of these
Sol.
S.NO.
1
2
3
4
[A] 0/M (Initial conc.)
–3
1.6 × 10
[B]0/M (Initial conc.)
rate / (M sec –1)
–2
5 × 10
5 × 10
4 × 10
–1
2 × 10
–1
8 × 10
1.6 × 10
–3
3.2 × 10
10
10
Rate law for reaction from above data is
(A) r = k[A]2 [B]2
(B) r = k[A]2[B]
2
(C) r = k [A][B]
(D) r = k[A][B]
Sol.
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–3
–3
3.2 × 10
–2
10
–3
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–3
–3
–3
Page # 110
7. For a reaction pA + qB
products, the rate law
expression is r = k[A]l[B]m, then :
(A) (p + q) < (1 + m)
(B) (p + q) > (1 + m)
(C) (p + q) may or may not be equal to (I + m)
(D) (p + q) = (1 + m)
Sol.
8. In the reaction : A + 2B
3C + D, which of the
following expression does not describe changes in the
conc. of various species as a function of time :
(A) {–d[C]/dt} = {3/2d[B]/dt}
(B) {3d[D]/dt} = {d[C]/dt}
(C) {3d[B]/dt} = {–2d[C]/dt}
(D) {2d[B]/dt} = {d[A]/dt}
Sol.
9. For the reaction, 2NO(g) + 2H2(g)
N2(g) + 2H2O
(g) the rate expression can be written in the following
ways : {d[N2]/dt} = k1[NO][H2] ; { d [ H 2 O ] / d t } =
k[NO][H2] ; {–d[NO]/dt} = k 1' , [NO][H2] ; {–d[H2]/
dt} = k"1[NO][H2]. The relationship between k, k1, k'1
and k"1 is :
(A) k = k1 = k'1 = k"1
(B) k = 2k1 = k'1 = k"1
(C) k = 2k'1 = k 1 k"1
(D) k = k1 = k'1 = k1 k"1
Sol.
CHEMICAL KINETICS
Sol.
11. A first order reaction is 50% completed in 20
minutes at 27°C and in 5 min at 47°C. The energy of
activation of the reaction is
(A) 43.85 kJ/mol
(B) 55.14 kJ/mol
(C) 11.97 kJ/mol
(D) 6.65 kJ/mol
Sol.
12. For the first order reaction A
B + C, carried
out at 27°C if 3.8 × 10–16% of the reactant molecules exists in the activated state, the Ea (activation energy) of the reaction is
(A) 12 kJ/mole
(B) 831.4 kJ/mole
(C)100 KJ/mole
(D) 88.57 kJ/mole
Sol.
13. The rate constant, the activation energy and
the Arrhenius parameter (A) of a chemical reaction at
25°C are 3.0 × 10–4 s–1, 104.4 kJ mol–1 and 6.0 × 10–4s–1
respectively. The value of the rate constant at T
is
(A) 2.0 × 1018 s–1
(B) 6.0 × 1014 s–1
(C) infinity
(D) 6 × 10–4 s–1
Sol.
10. At certain temperature, the half life period in the
thermal decomposition of a gaseous substance as follows :
P (mmHg)
500
250
t1/2 (in min)
235
950
Find the order of reaction [Given log (23.5) = 1.37 ;
log (95) = 1.97]
(A) 1
(B) 2
(C) 2.5
(D) 3
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CHEMICAL KINETICS
Page # 111
14. The following mechanism has been proposed for
the exothermic catalyzed complex reaction.
fast
A+B
I AB
k1
AB + I
k2
P+A
If k1 is much smaller than k2 . The most suitable
qualitative plot of potential energy (P.E.) versus
reaction coordinate for the above reaction.
16. Consider A
F.R.
B + heat, If activation energy
B.R.
for forward reaction is 100 kJ/mole then activation
energy for backward reaction and heat of reaction is
:
(A) 100, 200
(B) 80, 20
(C) 120, 220
(D) 140, 40
Sol.
AB+I A+P
(A)
A+B IAB
AB+I
A+P
(B)
Reaction coordinate
(C)
A+B
AB+I
A+P
IAB
Reaction coordinate
A+B
IAB
Reaction coordinate
(D)
A+B
IAB AB+I A+P
Reaction coordinate
17. In a reaction, the thershold energy is equal to :
(A) Activation energy + normal energy of reactants
(B) Activation energy – normal energy of reactants
(C) Activation energy
(D) Normal energy of reactants.
Sol.
Sol.
15. Choose the correct set of identifications.
(1)
(2)(3)
(4)
Reaction coordinate
(1)
(2)
(A) E for
E +S
ES
(B) Ea for
E +S
ES
(C) Ea for
ES
EP EP
(D) Ea for
E +S
ES
(E) E for
E+S
ES
Sol.
(3)
Ea for
ES
EPfor
E for
E+S ES
Ea for
E + P for
Ea for
ES
EPEP
EOv erall
for S
P
(4)
Eoverall
S
P
EP
Ea for
ES
EP
Eoverall
S
P
EP
Ea for
E + P for
E for
EP
E+P
Ea for
E+P
E overall
for S
P
E for
E+P
E overall
S
P
Ea for
EP
E+P
18. The first order rate constant k is related to temperature as log k = 15.0 – (106/T). Which of the following pair of value is correct ?
(A) A = 1015 and E = 1.9 × 104 KJ
(B) A = 10–15 and E = 40 KJ
(C) A = 1015 and E = 40 KJ
(D) A = 10–15 and E = 1.9 × 104 KJ
Sol.
19. When a graph between log K and 1/T is drawn a
straight line is obtained. The temperature at which
line cuts y-axis and x-axis.
(A) 0, Ea/2.303 R log A
(B) , Ea/(R ln A)
(C) 0, log A
(D) None of these
Sol.
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20. The rate constant, the activation energy and the
frequency factor of a chemical reaction at 25°C are 3.0 ×
10–2 s–1, 104.4 KJ mol–1 and 6.0 × 1014 s–1 respectively.
The value of the rate constant as T
is :
(A) 2.0 × 1018 s–1
(B) 6.0 × 1014 s–1
(C) infinity
(D) 3.6 × 1030 s–1
Sol.
21. The rate data for the net reaction at 25°C for
the reaction X + 2Y
3Z are given below :
[X0] [Y0 ]
Time required for [Z] to increase by
0.005 mol per litre.
0.01 0.01
72 sec
0.02 0.005 36 sec
0.02 0.01
18 sec
The intial rate (as given by Z) is :
(A) First order in both X and Y
(B) Second order in X and first order in Y
(C) First order in X and second order in Y
(D) None of the above
Sol.
CHEMICAL KINETICS
24. For a first order reaction, the concentration of
reactant :
(A) is independent of time
(B) varies linearly with time
(C) varies exponentially with time
(D) None
Sol.
25. Graph between conc. of the product and time of
the reaction A
B is of the type. Hence graph be–d[ A]
tween
and time will be of the type :
dt
1
x
time
[–d[A]/dt]
[–d[A]/dt]
(A)
(B)
time
time
[–d[A]/dt]
22. The rate of production of NH3 in N2 + 3H2 2NH3
is 3.4 kg min–1. The rate of consumption of H2 is :
(A) 5.1 kg min–1
(B) 0.01 kg sec–1
–1
(C) 0.6 kg hr
(D) None of these
Sol.
23. For a given reaction of first order it takes 20 min.
for the conc. to drop from 1.0 M to 0.60 M. The time
required for the conc. to drop from 0.60 M to 0.36 M
will be :
(A) more than 20 min
(B) 20 min
(C) less than 30 min
(D) cannot tell.
Sol.
[–d[A]/dt]
(C)
(D)
time
time
Sol.
26. Mathematical representation for t1/4 life for first
order reaction is over is given by :
(A) t1/4 = [(2.303)/(K)] log 4
(B) t1/4 = [(2.303/(K)] log 3
(C) t1/4 = [(2.303)/(K)] log(4/3)
(D) t1/4 = [(2.303)/(K)]log(3/4)
Sol.
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CHEMICAL KINETICS
Page # 113
27. For a reaction A
Products, the conc. of reactant C0, aC0, a 2C0, a 3C0............ after time interval 0,
t, 2t ............ where 'a' is constant. Then :
(A) reaction is of 1 st order and K = (1/t) ln a
(B) reaction is of 2nd order and K = (1/tC0)(1 – a)/a
(C) reaction is of 1st order and K = (1/t) ln (1/a)
(D) reaction is of zero order and K = {(1 – a)}C0/t
Sol.
/ Cu
28.
N2Cl
Cl + N2 Half-life
is independent of conc. of A. After 10 minutes volume
N2 gas is 10 L and after complete reaction 50 L. Hence
rate constant in min–1 :
(A) (2.303/10) log 5
(B) (2.303/10) log 1.25
(C) (2.303/10) log 2
(D) (2.303/10) log 4
Sol.
29. A reaction 2A + B
k
C + D is first order with
respect to A and 2nd order with respect to B. Initial
conc. (t = 0) of A is C0 while B is 2C0. If at t = 30
minutes the conc. of C is C0/4 then rate at t = 30
minutes is :
(A)
7C30k
27C30k
(B)
16
32
(C)
247C30k
64
(D)
Sol.
31. The decomposition of a gaseous substance (A)
to yield gaseous products (B), (C) follows first order
kinetics. If initially only (A) is present and 10 minutes
after the start of the reaction the pressure of (A) is
200 mm Hg and that of over all mixture is 300 mm Hg,
then the rate constant for 2A
B + 3C is
(A) (1/600) ln 1.25 sec–1 (B) (2.303/10) log 1.5 min–1
(C) (1/10) ln 1.25 sec–1 (D) None of these
Sol.
32. For a certain reaction of order 'n' the time for
half change t1/2 is given by t1/2 = [(2 – 2)/K] × C01/2,
where K is rate constant and C0 is the initial concentration. The value of n is :
(A) 1
(B) 2
(C) 0
(D) 0.5
Sol.
49 k C30
32
Sol.
30. In acidic medium the rate of reaction between
(BrO3)– and Br– ions is given by the expression, –
[d(BrO3–)/dt] = K[BrO3–][Br–][H +]2 It means :
(A) Rate constant of overall reaction is 4 sec–1
(B) Rate of reaction is independent of the conc. of acid
(C) The change in pH of the solution will not affect
the rate
(D) Doubling the conc. of H+ ions will increase the
reaction rate by 4 times.
33. Two first order reaction have half life in the ratio
3 : 2. Calculate the ratio of time intervals t1 : t2. The
time t1 and t2 are the time period for 25% and 75%
completion for the first and second reaction
respectively :
(A) 0.311 : 1
(B) 0.420 : 1
(C) 0.273 : 1
(D) 0.119 : 1
Sol.
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CHEMICAL KINETICS
34. A reaction proceeds in three stages, The first
stage is a slow second order reaction, the second
stage is fast and of first order, the third stage is fast
and is a third order reaction. The overall order of the
reaction is :
(A) First order
(B) Second order
(C) Third order
(D) Zero order.
Sol.
35. A reaction of the type A + 2B + C D occurs by
following mechanism
A+B
X rapid equilibrium
X+C
Y Slow
Y + B D Fast
What is the order of the reaction :
(A) 1
(B) 2
(C) 3
(D) Non determinable
Sol.
36. Following mechanism has been proposed for a reactions, 2A + B
D+E;
A+B
C + D .....(Slow) ;
A+C
E .......(fast).
The rate law expression for the reaction is :
(A) r = K[A]2
(B) r = K[A]2[B]
2
(C) R = K[A]
(D) r = K[A] [B]
Sol.
Sol.
38. A subtance undergoes first order decomposition.
The decomposition follows two parallel first order reactions as : and K1 = 1.26 × 10–4 sec–1; K2 = 3.8 ×
10–5 sec–1. The percentage distribution of B and C are
(A) 80% B and 20% C (B) 76.83% B and 23.17% C
(C) 90% B and 10% C (D) 60% B and 40% C
Sol.
39. For 2A
k1
k2
B + 3C, 2C
3D. Which of
k–1
the following is correct :
(A) d[C]/dt = 3k1[A]2 – 3k-1[B][C]3 – 2K 2[C]2
(B) d[B]/dt = k1 [A]2
(C) d[A]/dt = 2K-1[B][C]3 – 2K1 [B][C]3
(D) None.
Sol.
k
k
1
2
40. For 2A
B
3C. k1 =2×10-4 sec-1 and
-4
k2 = 3×10 l/mol-sec. {d[B]/dt} equal to:
(A) k1 [A] – k 2 [B]
(B) k1 [A]2 – k2 [B]
(C) k1 [A] – k 2 [B]2
(D) k1[A]2 – k2 [C]3
Sol.
41. For the reaction : A + 2B
37. For the reaction H2 + I2
k1
k2
2HI. The rate
law expression is :
(A) [(+1/2) d[HI]/dt] = k1 [H 2] [I2]
(B) [(+1/2) d[HI]/dt] = {k1 [HI]2 / k 2 [H2 ][I2]
(C) [(+1/2) d[HI]/dt] = k1 [H 2][I2] – k2[HI]2
(D) [(+1/2) d[HI]/dt] = k1 k2 [H2][I2]
AB2 ; the rate of
dx
= 1 × 105[A][B]2 – 1 × 104[AB2].
dt
The rate constants for forward and backward reactions are :
forward reaction is
(A) 1 × 105 L2 m–2 s–1, 1 × 104 sec
(B) 1 × 105 sec–1, 1 × 104 L2 m–2 s–1
(C) 1 × 105 L2 m–2 s–1, 1 × 104 sec–1
(D) 1 × 105 L m–1 s–1 , 1 × 104 sec
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CHEMICAL KINETICS
Page # 115
Sol.
44. For the reaction : 2NO + Br2
mechanism is given in two steps :
Fast
(1) NO + Br2
(2) NOBr2 + NO
2NOBr; the
NOBr2
2NOBr.
Slow
The rate expression for the reaction is :
42. At the point of intersection of the two curves
shown for the reaction
A
nB
(A) r = K[NO]2[Br2]
(B) r = K[NO][Br2]
(C) r = K[NO][Br2]2
Sol.
(D) r = K[NOBr2]
time
the concentration of B is given by :
(A)
nA 0
2
(B)
A0
n–1
(C)
nA 0
n 1
(D)
( n – 1)
A0
(n 1)
Sol.
45. For a gaseous reaction, the rate is expressed in
dP
dC
dn
in place of
or
, where C is condt
dt
dt
centration, n is number of moles and 'P' is pressure of
reactant. The three are related as :
terms of
(A)
dP
dt
1
RT dn
V dt
dP
(B) RT dt
43. For a hypothetical reaction aA + bB
the rate law is : rate = K[A]x[B]y. then :
Product,
(A) (a + b) = (x + y)
(B) (a + b) < (x + y)
(C) (a + b) > (x + y)
Sol.
(D) Any of these
(C)
dP
dt
dC
dt
1 dn
V dt
dn
dt
dC
dt
dC
dt
(D) None of these
Sol.
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CHEMICAL KINETICS
46. For a chemical reaction : A
Product, the rate
of disappearance of A is given by :
Sol.
–dC A
CA
K1
dt
1 K 2 C A . At low CA, the order of reaction
and rate constants are respectively :
K1
(A) I, K
2
(B) I, K1
K1
(C) II, K
2
K1
(D) II, K K
1
2
Sol.
49. How much faster would a reaction proceed at
25°C than at 0°C if the activation energy is 2 cal :
(A) 2 times
(B) 16 times
(C) 11 times
(D) Almost at same speed
Sol.
47. For a reversible reaction, A + B
C +D ;
H = – A kcal. If energy of activation for the forward
reaction is B kcal, the energy of activation for backward reaction in kcal is :
(A) – A + B (B) A + B
Sol.
(C) A – B
(D) – A – B
50. The temperature coefficient of reaction I is 2
and reaction II is 3. Both have same speed at 25°C
and show I order kinetics. The ratio of rates of reactions of these two at 75°C is :
(A) 7.6
(B) 5.6
(C) 6.6
(D) 8.6
Sol.
48. The reaction A (g)
B(g) + 2C (g) is a first
order reaction with rate constant 3465 × 10–6 s–1 .
Starting with 0.1 mole of A in 2 litre vessel, find the
concentration of A after 200 sec., when the reaction
is allowed to take place at constant pressure and
temperature.
(A) 0.05 M
(B) 0.025 M
(C) 0.0125 M
(D) None of these
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CHEMICAL KINETICS
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OBJECTIVE PROBLEMS (JEE ADVANCED)
EXERCISE – II
Sol.
Single correct
1. The reaction CH 3 – CH2 – NO2 + OH–
CH 3 – CH –
NO2 + H 2O obeys the rate law for pseudo first order
kinetics in the presence of a large excess of hydroxide ion. If 1% of nitro ethane undergoes reaction in
half a minute when the reactant concentration is 0.002
M, What is the pseudo first order rate constant ?
(A) 0.02 min–1
(B) 0.05 min–1
(C) 0.01 min–1
Sol.
(D) 0.04 min–1
K
4. A
B, t1/2 = 10 min
3A
C
Both reaction have same rate constant and each occurring following first order kinetics.
Choose the correct option for second reaction.
(A) t1/2 = 10/3 min (B) t1/2 = 30 min
(C) t1/2 = 10 min
(D) Data insufficient
Sol.
2. Decomposition of H2 O2 is a first order reaction. A
solution of H2O2 lebelled as "16.8 V" was left open.
Due to this, some H2O2 decomposed. To determine
the new volume strength after 2.303 hours, 20 mL of
this solution was diluted to 100 mL. 25 mL of this
diluted solution was titrated against 37.5 mL of 0.02
M KMnO4 solution under acidic conditions [Given : STP
is 1 atm and 273 K] Calculate the rate constant (in
5. The gas phase decomposition (in closed container)
hr–1) for decomposition of H2O2.
(A) 0.6
Sol.
(B) 0.5
(C) 0.4
(D) 0.8
2A(g)
4B(g) + C(g)
Follows the first order rate law. At a given temperature specific reaction rate is 7.5 × 10–3 s–1. The
initial pressure of A is 0.1 atm. Calculate the time of
decomposition of A so that the total pressure becomes 0.15 atm. (log 1.4925 = 0.1739).
–
3. The reaction A
OH
B, obeys the rate law for
(A) 66.6 sec
(B) 22.24 sec
(C) 53.4 sec
(D) None of these
Sol. C
pseudo first order kinetics in the presence of a large
excess of hydroxide ion. If 90% of A undergoes reaction in half a minute when the reactant concentration
is 0.002 M, What is the pseudo first order rate constant in min–1
(A) 4.61
(B) 2.61
(C) 3.61
(D) 5.61
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CHEMICAL KINETICS
6. Which graph represents zero order reaction
[A(g)
(A)
B(g)] :
d [B ]
(B) dt
[B]
t
t
8. For the first order decomposition of SO2Cl 2(g),
SO2Cl 2 (g)
SO2 (g) + Cl2 (g)
a graph of log (a0 – x) vs t is shown in figure. What
is the rate constant (sec–1 )?
Time (min)
2
4 6 8 10
(0,0)
|
| | |
|
-1–
-2–
-3–
t1/2
(C)
(D)
(B) 4.6 × 10–1
(D) 1.15 × 10–2
(A) 0.2
(C) 7.7 × 10–3
Sol.
t3/4
[A]0
[A]0
Sol.
7. If decomposition reaction A (g)
B (g) follows
first order kinetics then the graph of rate of formation
(R) of B against time t will be
(A)
9. The variation of concentration of A with time in
two experiments starting with two different initial
concentration of A is given in the following graph. The
reaction is represented as A(aq)
B(aq). What is
the rate of reaction (M/min) when concentration of A
in aqueous solution was 1.8 M?
(B)
1.5
1.2
1
(C)
(D)
0.8
0.6
Sol.
(A) 0.08 M min–1
(C) 0.13 M min–1
Sol.
Experiment-1
Experiment-2
5
10
15
20
time(min.)
(B) 0.036 M min–1
(D) 1 M min–1
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CHEMICAL KINETICS
Page # 119
10. SO3 gas is entering the environment at a constant
rate of 6.93 × 10–6 gm/L/day due to the emission of
polluting gases from thermal power plant, but at the
same time it is decomposing & following first order
kinetics with half life of 100 days.
Based on above i nformation select th e true
statement(s).
(A) Concentration of SO3 in Kota is 1.25 × 10–5 M
(Assume SO3 present in air reaches steady state)
(B) If 103 L of air is passed through 1 L pure water
(assuming all SO3 to be dissolved in it) & resulting
solution is titrated against 1 N NaOH solution, 15 ml is
required to reach end point.
(C) An industry is manufacturing H2SO4 at the rate of
980 kg per day with the use of SO3 in air it should
use 8 × 105 Litre air /day.
(D) If SO3 emission is stopped then after 1000 days
its concentrations will reduce to ~ 1.2 ×10–8 M.
Sol.
11. For the reaction A
is
Sol.
12. Consider the reaction,
B
A
C
A, B and C all are optically active compound . If optical
rotation per unit concentration of A, B and C are 60°,
–72°, 42° and initial concentration of A is 2 M then
select write statement(s).
(A) Solution will be optically active and dextro after
very long time
(B) Solution will be optically active and levo after
very long time
(C) Half life of reaction is 15 min
(D) After 75% conversion of A into B and C angle of
rotation of solution will be 36°.
Sol.
B, the rate law expression
d[ A ]
= k [A]1/2. If initial concentration of [A] is
dt
[A]0 , then
(A) The intege rated rate exp ress ion i s k =
2 1/ 2
(A
t 0
A1 / 2 )
(B) The graph of
Directions : Read the following questions and
choose
A Vs t will be
K
(C) The half life period t 1 / 2 =
2[ A ]10/ 2
(D) The time taken for 75% completion of reaction
t 3/ 4 =
[ A ]0
REASONING TYPE
(A) If both the statements are true and statement-2 is the correct explanation of satement-1.
(B) If both the statements are true and statement-2 is not the correct expla nation of
satement-1.
(C) If statement-1 is True and statement-2 is
False.
(D) If statement-1 is False and statement-2 is
True.
k
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Assertion & Reasoning type questions
13. Statement-1 :
A fractional order reaction
must be a complex reaction.
Statement-2 :
Fractional order of RDS equals to
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
14. Statement-1 :
The time of completion of
reactions of type A
product (order <1) may be
determined.
Statement-2 :
Reactions with order 1 are either
too slow or too fast and hence the time of
completion can not be determined.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
15. Statement-1 :
Temperature coefficient of an
one step reaction may be negative.
Statement-2 :
The rate of reaction having
negative order with respect to a reactant decreases
with the increase in concentration of the reactant.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
16. Statement-1 :
Th e overal l rate o f a
reversible reaction may decrease with the increase in
temperature.
Statement-2 :
When the activation energy of
CHEMICAL KINETICS
forward reaction is less than that of backward
reaction, then the increase in the rate of backward
reaction is more than that of forward reaction on
increasing the temperature.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
17. Statement-1 :
In a reversible endothermic
reaction, Eact of forward reaction is higher than that
of backward reaction
Statement-2 :
The threshold energy of forward
reaction is more than that of backward reaction
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
18. Statement-1 :
A catal ys t prov ides an
alternative path to the reaction in which conversion
of reactants into products takes place quickly
Statement-2 :
The catalyst forms an activated
complex of lower potential energy, with the
reactants by which more number of molecules are able
to cross the barrier per unit of time.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
19. Statement-1 : Rate of a chemical reaction increases as the temperature is increased.
Statement-2 : As the temperature is increased fraction of molecules occupying ET or more increases.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
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CHEMICAL KINETICS
Page # 121
Sol.
22. Ozone decomposes according to the equation
2O3(g)
3O2(g) Mechanism of the reaction is
Step I :
O3(g)
O2(g) + O(g)
Step II :
O3(g) + O(g)
2O2(g)
(fast)
(slow)
Which of the following is correct ?
(A) for step I, molecularity is 2 (B) for step II,
molecularity is 1
(C) step II is rate determining step
(D) Rate law expression for the overall reaction is
–
20. Statement-1 : For the bimolecular reaction to
react, reactant must collide with each other.
d
[O 3 ] k[ O 3 ]2 [ O2 ] –1
dt
Sol.
Statement-2 : Bond breaking and formation occurs
during the collision.
(A) (A)
(B) (B)
(C) (C)
(D) (D)
Sol.
23. Consider the following case of competing 1st order
reactions
After the start of the reaction at t = 0, with only P,
concentration
of Q is equal to R at all times. The time in which all
Q
k
1
21. Statement-1 : Catalyst does not change the H
value of the reaction.
Statement-2 : Catalyst are generally added in very
small quantities and not stoichiometrically.
(A) (A)
Sol.
(B) (B)
(C) (C)
(D) (D)
the three
P
k2 R
concentration will be equal is given by
(A) t =
(C) t
2.303
log10 3
2k1
2.303
log10 2
3k1
(B) t
2.303
log10 3
2k 2
(D) t
2.303
log10 2
3k 2
Sol.
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24. A substance undergoes first order decomposition.
The decomposition follows two parallel first order
reactions as
k1
B k1=1.26 × 10–4 s–1
A
k2=3.8 × 10–5 s–1
k2 C
The % distribution of B and C is ................ and
.................. respectively.
(A) 76.83% B
(B) 23.17% C
(C) 23.17%
(D) 76.83%C
Sol.
25. Which of the following concepts are correct
(A) If for, A
B
H = +q cal
then for, B
A H = – q cal
(B) If for, A
B equilibrium constant = K
1
then for, B
A equilibrium constant =
K
(C) If for, A
B rate constant = k
then for, B
A rate constant k
(D) If for, A
B energy of activation = E
then for, B
A energy of activation = – E
Sol.
26. The basic theory behind Arrhenious equation is
that
(A) the activation energy and pre exponential factor
are always temperature independent.
(B) the number of effective collisions is proportional
to the number of molecules above a certain threshold
energy.
(C) as the temperature increases, so does the number
of molecules with energies exceeding threshold energy.
(D) rate constant is a function of temperature.
Sol.
CHEMICAL KINETICS
27. Which of the following statements are correct
about half life period ?
(A) It is proportional to initial concentration for a zero
order reaction.
(B) Average life = 1.44 times half life for a first order
reaction.
(C) Time of 75% reaction is thrice of half life period in
second order reaction.
(D) 99.9% reaction which is first order, takes place in 100
minutes if the rate constant of the reaction is 0.0693 min–1.
Sol.
28. For a reaction : 2A + 2B
products, the rate
law expression is r = k[A]2 [B]. Which of the following
is/are correct?
(A) The reaction is first order w.r.t. B
(B) The reaction is of second order w.r.t. A
(C) The reaction is of third order, overall
(D) Slowest step of the reaction is given as A + B AB
Sol.
29.
Et
Et
Cl + OH–
H
Me
H
Et
OH,
OH
H
Me
Me
(I)
(II)
Which of the following statements are correct ?
(A) It is SN 1 if (I) or (II) is formed.
(B) It is SN1 if equimolar mixture of (I) and (II) is
formed.
(C) It is SN2 if (I) or (II) is formed.
(D) It is SN2 if (II) is formed.
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CHEMICAL KINETICS
Sol.
30. Rate constant k varies with temperature by
2000
. We can conclude
T
(A) pre exponential factor A is 105
(B) Ea is 2000 k cal
(C) Ea is 9.212 k cal
(D) pre exponential factor A is 5
Sol.
equation log10k(min–1) = 5 –
31.
A reaction is catalysed by H+ ion. In presence
of HA, rate constant is 2 × 10–3 min–1 and in presence
of HB rate constant is 1 × 10–3 min–1. HA and HB being
strong acids, we may conclude that
(A) HB is stronger acid than HA
(B) HA is stronger acid than HB.
(C) relative strength HA and HB is 2.
(D) HA is weaker than HB and relative strength is 0.5.
Sol.
32. The potential energy diagram for a reaction R
P is given below
Which of the following is true ?
z
(A) Ea (forward) = y
(B) Ea (backward) = z
y
x
(C) ET = x + y
R
(D) Hr = x
Sol.
Page # 123
33. An organic compound A decomposes by following
two parallel first order mechanisms :
k1
B
k1 1
= , k = 0.693 hr –1
A
;
1
k
k2
2 2
C
Select the correct statement(s)
(A) If three moles of A are completely decomposed
then 2 moles of B and 1 mole of C will be formed.
(B) If three moles of A are completely decomposed
then 1 moles of B and 2 mole of C will be formed.
(C) half life for the decomposition of A is 20 min
(D) half life for the decomposition of B is 0.33 min
Sol.
34. Half life for which of the following varies with
initial concentration [A0]
(A) zero order
(B) first order
(C) second order
(D) third order
Sol.
35. In which of the following, Ea for backward
reaction is greater than Ea for forward reaction ?
(A) a
Ea 50 kcal
(B) a
b
H
10 kcal
Ea 60 kcal
b H
20kcal
(C) a
Ea 50 kcal
b H
10 kcal
(D) a
Ea 60 kcal
b H
20kcal
Sol.
COMPREHENSION - 1
Oxidation of metals is generally a slow electrochemical reaction involving many steps. These steps involve electron transfer reactions. A particular type of
oxidation involve overall first order kinetics with re: 0744-2209671, 08003899588 | url : www.motioniitjee.com,
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Page # 124
CHEMICAL KINETICS
spect to fraction of unoxidised metal (1 – f) surface
thickness relative to maximum thickness (T) of oxidised
surface, when metal surface is exposed to air for considerable period of time.
200 hrs
0
t
39. Number of moles of B are
(A) 2
(B) 1
(C) 0.666
Sol.
(D) 0.333
COMPREHENSION - 3
–3
ln (1 – f)
df
x
k(1 – f ) , where f = , x = thickness
Rate law :
dt
T
of oxide film at time 't' & T = thickness of oxide film at
t=
A graph of ln (1 – f) vs t is shown in the
adjacent figure.
36. The time taken for thickness to grow 50% of 'T' is
(A) 23.1 hrs (B) 46.2 hrs (C) 100 hrs (D) 92.4 hrs
Sol.
The rate of a reaction increases significantly with increase in temperature. Generally, rates of reaction
are doubled
for every 10º rise in temperature.
Temperature coefficient gives us an idea about the
change in rate of a reaction
for
ev ery
10 º
change in temperature.
Temperature coefficient
Rate constant of ( T 10)º C
Rate constant at Tº C
Arrhenious gave an equation which describes rate constant k as a function of temperature is k A Ea / RT
where k is a rate constant
A is frequency factor or pre exponential factor
Ea is activation energy
T is temperature in kelvin and
R is universal gas constant
37. The exponential variation of 'f' with t(hrs) is given by
(A) [1 – e –3t / 200 ]
(B) e–3 t/ 200 – 1
(C) e –3 t/ 200
Sol.
(D) e3 t/ 200
Equation when expressed in logarithmic form becomes
log10 k
log10 A
Ea
2. 303RT
40. For a reaction Ea = 0 and k = 3.2 × 105 s–1 at 325 K.
The value of k at 335 K would be
(A) 3.2 × 105 s–1
(B) 6.4 × 108 s–1
(C) 12.8 × 108 s–1
(D) 25.6 × 108 s–1
Sol.
COMPREHENSION - 2
k1
2B
k2
C
For a hypothetical elementary reaction A
k1
where k
2
1
2
Initially only 2 moles of A are present.
38. The total number of moles of A, B & C at the end
of 50% reaction are
(A) 2
(B) 3
(C) 5
(D) None
Sol.
41. For which of the following reactions k310/k300 would
be maximum ?
(A) P Q
R ; Ea
10kJ
(B) E F
D; Ea
21kJ
(C) A B
C; Ea
10.5 kJ
(D) L M
N; E a
5 kJ
Sol.
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CHEMICAL KINETICS
Page # 125
42. Activation energies of two reaction are Ea and E a
with Ea > E a. If the temperature of the reacting systems is increased from T1 to T2 (k are rate constants
at higher temperature).
k1
(A)
k1
k2
k2
k1
(B)
k1
k1
k1
k2
k2
(D)
(C)
k1
k1
k2
k2
44. The rate of formation of SO3 in the following reaction : 2SO2 + O2
2SO3 is 10 g sec–1. The rate of
disappearance of O2 will be :
(A) 5 g sec–1
(B) 100 g sec–1
(C) 20 g sec
(D) 2 g sec–1
Sol.
2k 2
k2
Sol.
45. For the reaction : aA
log
– dA
dt
(A) 3.98
Sol.
log
dB
dt
bB ;
0.6 , then a : b is :
(B) 2.18
(C) 1.48
(D) 0
43. For the reactions, following data is given
P
Q
k 1 1015 exp .
2000
T
C
D
k2
1014 exp.
1000
T
Temperature at which k1 = k2 is
(A) 434.22 K (B) 1000 K (C) 2000 K (D) 868.44 K
Sol.
46. For a reaction, 2ND3
– d[ND 3 ]
dt
d[D2 ]
dt
COMPREHENSION - 4
The rate and mechanism of chemical reactions are
studied in chemical kinetics. The elementary reactions are single step reactions having no mechanism.
The order of reaction and molecularity are same for
elementary reactions. The rate of forward reaction
aA + bB
cC + dD is given as :
rate =
=
d[N2 ]
dt
=
K 2 [ND 3 ]
;
K 3 [ND 3 ] , then :
(A) K1 = K 2 = K 3
(C) K1 = 2K 2 = K 3
(B) 3K1 = 6K2 = 2K3
(D) K1 = K2 = 2K3
Sol.
dx
1 d[A]
1 d[B] 1 d[C] 1 d[D]
=–
=–
=
=
or rate =
dt
a dt
b dt c dt d dt
K[A]a[B]b. In case of reversible reactions net rate expression can be written as : rate = K1 [A]a[B]b –
K2[C]c[D]d. At equilibrium, rate = 0. The constants K,
K1, K2, are rate constants of respective reaction. In
case of reactions governed by two or more steps reactions mechanism, the rate is given by the solwest
step of mechanism.
K 1 [ND 3 ] ;
N2 + 3D2 ;
COMPREHENSION - 5
The rate of a reaction
dx
dt
varies with nature, physi-
cal state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate
constant (K) varies with temperature and catalyst
only. The rate constant K is given as K Ae–Ea /RT
where A is Arrhenius parameter or pre-exponential fac-
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CHEMICAL KINETICS
tor and Ea is energy of activation. The minimum energy required for a reaction is called threshold energy
and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation.
47. For a reaction, A
MATCH THE FOLLOWING
50. Match the column
Column - I
(A) First order reaction
B; if
125
.
10 4
K, the Arrhenius paT
rameter and energy of activation for the reaction are
log10 K(sec–1) = 14 –
(A) 1014 sec–1, 239.34 kJ
(B) 14,57.6 kcal
(C) 1014 sec–1, 23.93 kJ
Sol.
(D) 1014 sec, 5.76 kcal
(B) Second order reaction
(C) Zero order reaction
(D) Fractional (+ ve)
order reaction
Column - II
(P) Rate constant
increases on increasing
the concentration
(Q) Half life depends on
the initial concentration
(R) Reaction must be
complex
(S) Half life decreases on
increasing the temperature
(T) The plot of
concentration of
reactant verses time will
be a rectangular hyperbola
Sol.
48. At what conditions exponential factor is 1 for a
reaction :
(A) Infinite temperature
(B) Free radical combination
(C) Ea = 0
(D) All of these
Sol.
49. For an endothermic reaction, which one is true if
H is heat of reaction and Ea is energy of activation :
(A) Ea > H
(B) Ea < H
(C) Ea >< H
Sol.
(D) Ea = 0
51. Consider the following energy diagram for the reaction.
A2 + B2
2AB
Energy
(j mol–1)
Column I
(A) Ea(f)
(B) Ea(b)
(C) Hr
(D) ET
50
40
30
20
10
Column II
(P) –10 kJ mol–1
(Q) 40 kJ mol–1
(R) 30 kJ mol –1
(S) 50 kJ mol –1
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CHEMICAL KINETICS
Page # 127
Column I
(Graphs reaction A
Sol.
52.
(B)
(C)
(D)
d[ B]
vs
dt
d[ A ]
for first order
dt
[A] vs t for first order
[B] vs t for first order
[A] vs t
for zero order
Column II
(P)
(Q)
(R)
(S)
(A)
(B)
(C)
(D)
Column II
(Co-ordinates)
(P)
ln [A] (y-axis), t (x-axis) (order = 1)
(Q)
t1/2 (y-axis), [A0] (x-axis) (order = 1)
(R)
r (y-axis), t (x-axis) (order > 0)
(S)
r (y-axis), t (x-axis) (order = 0)
(T)
t1/2 (y-axis), [A0] (x-axis) (order > 1)
For the reaction of type A(g)
2B(g)
Column-I contains four entries and columnII contains four entries. Entry of column-I are
to be matched with ONLY ONE ENTRY of
column-II
Column I
(A)
Products)
(U)
1
(y-axis), t (x-axis) (order = 2)
[A ]
(V)
r (y-axis), [A] (x-axis) (order = 1)
Sol.
INTEGER TYPE
54. The complex [CO(NH3)5F]2+ reacts with water
Sol.
according to the equation.
[ Co(NH3 ) 5 F] 2
H2 O
[ Co(NH3 ) 5 (H2 O)] 3
F–
The rate of the reaction = rate const. × [complex]a
× [H+]b. The reaction is acid catalysed i.e. [H +] does
not ch ange during the reaction. Thus rate =
k[Complex]a where k' = k [H+]b, calculate 'a' and 'b'
given the following data at 25°C.
[Complex] M
[H +] M
0.1
0.01
1
2
0.02
0.5
1
0.2
53.
T1/ 2 hr
Column-I and column-II. Entry of column-I
are to be matched with ONE OR MORE THAN
ONE ENTRIES of column-II and vice versa.
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T3 /4 hr
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CHEMICAL KINETICS
Sol.
57. For a 1st order reaction (gaseous) (cont. V, T)
aA
(b – 1) B + C (with b > a ) the pressure of the
b
system increased by 50 a 1 % in a time of 10 min.
The half life of the reaction is therefore (in min.).
Sol.
55. An optically active compound A upon acid
catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The
observed rotation of the mixture after 20 min was 5°
while after completion of the reaction it was –20°. If
optical rotation per mole of A, B & C are 60°, 40° & –
80°. Calculate half life of the reaction.
Sol.
58. The rate of decomposition of NH3(g) at 10 atm
on platinum surface is zero order. What is rate of
formation (in M min–1) of H2(g), if rate constant of
reaction 2NH3(g) N2 (g)+3H3 (g) is 2.0 M min–1 ?
Sol.
56. Consider the following first order decomposition
process:
An
t
nA
time
Here, “t” corresponds to the time at which
reactant is decomposed. The value of “n” is
Sol.
1
6
th
of
59. 5A
Product
In above reaction, half-life period is directly proportional
to initial concentration of reactant. The initial
rate of reaction is 400 mol lit–1 min–1 .
Calculate the half-life period (in sec) when initial
concentration of reactant is 200 mol lit–1 .
Sol.
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Page # 129
60. In an elementary reaction A(g) +2B(g)
C(g)
the initial pressure of A and B are PA=0.40 atm and PB
=0.60 atm respectively. After time T, if pressure of
C is observed 0.1 atm, then find the value of
ri(initial rate of reaction)
rt (rate of reaction after time t)
Sol.
63. For any acid catalysed reaction, A
H
B
Half-life period is independent of concentration of A
at given pH. At same concentration of A half-life
time is 10 min at pH=2 and half-life time is 100 min at
pH=3. If the rate law expression of reaction is
r=k[A]x[H+]y then calculate the value of (x+y).
Sol.
61. Carbon monoxide reacts with O2 to form CO2:
2CO(g)+O2(g)
2CO2(g) information on this reaction
is
given in the table below.
[CO] mol/L [O2] mol/L
0.02
0.04
0.02
0.02
0.02
0.04
Rate of reaction (mol/L.min)
–5
4×10
–4
1.6×10
–5
8×10
What is the value for the rate constant for the reaction
in proper related unit ?
Sol.
64. For a reaction, A
B equilibrium constant is
1.66 and kforward= 0.166 hr–1.
Calculate the time (in hours) when concentration of
B is 80% of its equilibrium concentration.
(Given : ln 25=3.20)
Sol.
62. Half- life f or the z ero order reac tion ,
A(g) B(g)+C(g) and half-life for the first order
reaction X(g)
Y (g)+Z (g) i s equ al. If
completion time for the zero order reaction is 13.86
min, then calculate the rate
constant (in hr–
1
) for the reaction X(g) Y(g)+Z(g).
Sol.
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CHEMICAL KINETICS
SUBJECTIVE PROBLEMS (JEE ADVANCED)
EXERCISE – III
Q.1 For the reaction
2NH3 N2 + 3H2
a curve is plotted between [NH3] is time as shown
Q.3 Dinitropentaoxide decomposes as follows :
N2O5 (g)
Given that
0.6
0.2
5
10 15
Time (sec.)
k1[N2 O 5 ]
20
Calculate
(a) rate of disappearance between 5 to 10 sec
(b) rate of disappearance between 10 to 20 sec
Sol.
Q.2 For the reaction 3BrO–
BrO3– 2Br – in an alkaline aquesous solution, the value of the second order
(in BrO–) rate constant at 80°C in the rate law for
–
[BrO ]
was found to be 0.056 L mol –1s–1. What is
t
the rate of constant when the rate law is written for
–
[BrO –3 ]
, (b)
t
– d[N2 O 5 ]
dt
1
O 2 ( g)
2
d[NO2 ]
d[ O2 ]
k2 [N2 O 5 ]
k 3 [N2 O5 ]
dt
dt
What is the relation between k1, k2 and k3 ?
Sol.
0.4
(a)
2NO2(g) +
[Br – ]
?
t
Q.4 The reaction 2A + B + C
D + E is found to be
first order in A second order in B and zero order in C.
(i) Give the rate law for the reaction in the form of
differential equation.
(ii) What is the effect in rate of increasing concentrations of A, B, and C two times ?
Sol.
Q.5 For the elementary reaction 2A + B2 2AB. Calculate how much the rate of reaction will change if
the volume of the vessel is reduced to one third of its
original volume?
Sol.
Sol.
Q.6 Ammonia and oxygen reacts at higher temperatures as
4NH3(g) + 5O2(g) 4NO(g) + 6H2O (g)
In an experiment, the concentration of NO increases
by 1.08 × 10–2 mol litre–1 in 3 seconds. Calculate.
(i) rate of reaction.
(ii) rate of disappearance of ammonia
(iii) rate of formation of water.
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Sol.
Page # 131
Sol.
Q.7 In the following reaction 2H2O2 2H 2O + O2
rate of formation of O2 is 3.6 M min–1.
(a) What is rate of formation of H2O ?
(b) What is rate of disappearance of H2O2 ?
Sol.
Q.8 The reaction A(g) + 2B(g)
C(g) + D (g) is an
elementary process. In an experiment, the initial partial pressure of A & B are PA = 0.6 and PB = 0.8 atm, if
PC = 0.2 atm then calculate the ratio of rate of reaction relative to initial rate.
Sol.
ZERO ORDER
Q.9 In the following reaction, rate constant is
1.2 × 10–2 Ms–1 A
B. What is concentration of B
after 10 and 20 min., if we start with 10 M of A.
Q.11 The rate constant for a zero order reaction is
2 × 10–2 mol L–1 sec–1, if the concentration of the
reactant after 25 sec is 0.25 M, calculate the initial
concentration.
Sol.
Q.12 A drop of solution (volume 0.10 ml) contains 6
× 10–6 mole of H +, if the rate constant of disappearance of H+ is 1 × 107 mole litre–1 sec–1. How long
would it take for H+ in drop to disappear ?
Sol.
Q.13 A certain substance A is mixed with an equimolar quantity of substance B. At the end of an hour A is
75% reacted. Calculate the time when A is 10%
unreacted. (Given : order of reaction is zero)
Sol.
Sol.
FIRST ORDER
Q.14 A first order reaction is 75% completed in 72
min. How long time will it take for
(i) 50% completion
(ii) 87.5% completion
Sol.
Q.10 For the followin g data for the reaction
A
products. Calculate the value of k.
Time (min.)
[A]
0.0
0.10 M
1.0
0.09 M
2.0
0.08 M
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Q.15 A first order reaction is 20% complete in 10
min. Calculate (i) the specific rate constant, (ii) the
time taken for the reactions to go to 75% completion.
Sol.
CHEMICAL KINETICS
Q.19 A viral preparation was inactivated in a chemical bath. The inactivation process was found to be
first order in virus concentration. At the beginning of
the experiment 2.0% of the virus was found to be
inactivated per minute. Evaluate k for inactivation process.
Sol.
Q.16 Show that in case of unimolecular reaction, the
time required for 99.9% of the reaction of take place in
ten times that required for half of the reaction.
Sol.
Q.20 Consider the reaction : A B + C. Initial concentration of A is 1 M. 20 minutes time is required for
d[B]
k[ A ], then calcucompletion of 20% reaction. If
dt
late half life (t1/2) of reaction.
Sol.
Q.17 A first order reaction has a rate constant is
1.5 × 10–3 sec –1. How long will 5.0 g of this reactant
take to reduce to 1.25 g.
Sol.
Q.18 A drug is known to be ineffective after it has
decomposed 30%. The original concentration of a
sample was 500 units/ml. When analyzed 20 months
later, the concentration was found to be 420 units/
ml. Assuming that decomposition is of I order, what
will be the expiry time of the drug?
Sol.
Q.21 The reaction SO2Cl 2(g)
SO2(g) + Cl2(g) is a
first order gas reaction with k = 2.2 × 10–5 sec–1 at
320° C. What % of SO2Cl2 is decomposed on heating
this gas for 90 min.
Sol.
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Page # 133
ORDER OF REACTION & RATE LAW
Sol.
Q.22 At 800° C the rate of reaction
2NO + H 2 N2 + H 2O
Changes with the concentration of NO and H2 are
1 d[NO]
in M sec–1
2 dt
(i) 1.5 × 10–4
4 × 10–3
4.4 × 10–4
(ii) 1.5 × 10–4
2 × 10–3
2.2 × 10–4
(iii) 3.0 × 10–4 2 × 10–3
8.8 × 10–4
(a) What is the order of this reaction ?
(b) What is the rate equation for the reaction ?
(c) What is the rate when
[H 2] = 1.5 × 10–3 M and [NO] = 1.1 × 10–3 M ?
Sol.
[NO] in M
[H2] in M
–
Q.23 The data below are for the reaction if NO and
Cl2 to form NOCl at 295 K
Concentration of Cl2[M] Concentration of NO
Initial
Rate (Ms–1)
0.05
0.05
1 × 10–3
0.15
0.05
3 × 10–3
0.05
0.15
9 × 10–3
(a) What is the order w.r.t NO and Cl2 in the reaction.
(b) Write the rate expression
(c) Calculate the rate constant
(d) Determine the reaction rate when concentration
of Cl2 and NO are 0.2 M & 0.4 M respectively.
Sol.
Q.25 The following data are for the reaction A + B
products :
Conc. A
Conc. B
Initial Rate
(M)
(M)
(mol L–1 s–1)
0.1
0.1
4.0 × 10–4
0.2
0.2
1.6 × 10–3
0.5
0.1
2.0 × 10–3
0.5
0.5
1.0 × 10–2
(i) What is the order with respect to A and B for the
reaction ?
(ii) Calculate the rate constant.
(iii) Determine the reaction rate when the concentrations of A and B are 0.2 M and 0.35M, respectively.
Sol.
Q.26 The pressure of a gas decomposing at the surface of a solid catalyst has been measured different
times and the results are given below
t (sec)
0
100
200
300
Pr. (Pascal)4 × 103 3.5 × 103
3 × 103 2.5 × 103
Determine the order of reaction, its rate constant.
Sol.
Q.24 The catalytic decomposition of N2O by gold at
900° C and at an initial pressure of 200mm is 50%
complete in 53 minutes and 73% complete in 100 minutes.
(i) What is the order of the reaction ?
(ii) Calculate the velocity constant.
(iii) How much of N2O will decompose in 100 min. at
the same temperature but at initial pressure of
600 mm?
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CHEMICAL KINETICS
Q.27 The half life period of decomposition of a compound is 50 minutes. If the initial concentration is
halved, the half life period is reduced to 25 minutes.
What is the order of reaction ?
Sol.
Sol.
Q.31 Show that in case of a first order reaction, the
time required for 93.75% of the reaction to take place
is four times that required for half of the reaction.
Sol.
Q.28 At 600°C, acetone (CH3COCH3) decomposes to
ketene (CH2 = C = O) and various hydrocarbons. Given
the initial rate data in the table :
(a) What is the order ?
(b) Write rate law
(c) Calculate rate constant
(d) Calculate the rate of decomposition when the
acetone concentration is 1.8 × 10–3 M
Experiment Initial [CH3COCH3]
Rate M s –1
1.
6.0 × 10–3 M
5.2 × 10–5
2.
9.0 × 10–3 M
7.8 × 10–5
–3
3.
1.8 × 10 M
?
Sol.
Q.32 The half time of the first order decomposition
of nitramide is 2.1 hour at 15°C.
NH 2NO2(aq.) N2 O (g) + H2O (l)
If 6.2 g of NH2NO2 is allowed to decompose, calculate
(i) time taken for NH2NO2 to decompose 99%, and (ii)
volume of dry N2O produced at this point, measured
at 1 atm & 273 K.
Sol.
HALF LIFE
Q.29 The half life period of a first order reaction is
50 min. In what time will it go to 90% completion ?
Sol.
Q.30 A first order reaction has k = 1.5 × 10–6 per
second at 200°C. If the reaction is allowed to run for
10 hrs., what percentage of the initial concentration
would have changed into the product ? What is the
half life of this reaction ?
Q.33 A flask contains a mixture of compounds A and
B. Both compounds decompose by first-order kinetics. The half-lives are 54.0 min for A and 18.0 min. for
B. If the concentartions of A and B are equal initially,
how long will it take for the concentration of A to be
four times that of B ?
Sol.
Q.34 Two substances A (t1/2 = 5 mins) and B (t1/2 =
15 mins) follow first order kinetics are taken in such a
way that initially [A] = 4[B]. Calculate the time after
which the concentration of both the substance will
be equal.
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CHEMICAL KINETICS
Page # 135
Sol.
Sol.
CONCENTRATION REPLACED BY OTHER QUANTITIES IN FIRST ORDER INTEGRATED RATE LAW
Q.35
In this case we have
A
B
Time
t
Total pressure of A + B + C P2
Find k.
Sol.
Q.36 A
B+C
Time
Total pressure of (B + C)
Find k.
Sol.
+
Q.39 S
G+F
Time
t
Rotation of Glucose & Fructose rt
Find k.
Sol.
r
C
P3
Q.40 At 27°C it was observed during a reaction of
hydrogenation that the pressure of hydrogen gas decreases from 2 atmosphere to 1.1 atmosphere in 75
minutes. Calculate the rate of reaction (in M sec–1)
and rate of reaction in terms of pressure.
Sol.
t
P2
P3
Q.37 A
B+C
Time
0
t
Volume of reagent
V1
V2
The reagent reacts with A, B and C. Find k.
Sol.
Q.41 At 100° C the gaseous reaction A
2B + C
was observed to be of first order. On starting with
pure A it is found that at the end of 10 minutes the
total pressure of system is 176 mm. Hg and after a
long time 270 mm Hg. From these data find (a) initial
pressure of A (b) the pressure of A at the end of 10
minutes (c) the specific rate of reaction and (d) the
half life period of the reaction ?
Sol.
Q.38 A
2B + 3C
Time
t
Volume of reagent
v2
v3
Reagent reacts with all A, B and C. Find K.
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CHEMICAL KINETICS
3
H2 ( g) was
2
followed at constant volume at 310°C by measuring
the gas pressure at intervals Show from the following
figures that reaction is of first order.
Time (in hrs)
0
5
7.5 10
Total pressure (in mm) 758
827
856
882
Sol.
Q.42
The reaction AsH3 (g)
As(s) +
Q.43 The decomposition of N2O5 according to the
equation 2N2O5 (g) 4 NO2(g) + O2 (g) is a first order
reaction. After 30 min. from start of decomposition in
a closed vessel the total pressure developed in found
to be 284.5 mm Hg. On complete decomposition, the
toal pressure is 584.5 mm Hg. Calculate the rate constant of the reaction.
Sol.
Q.44 The thermal decomposition of dimethyl ether
as measured by finding the increase in pressure of the
reaction.
(CH3)2 O(g) CH4(g) + H2(g) + CO (g)
At 500° C is as follows :
Time (sec.)
390
1195
3155
Pre. increase (mm Hg) 96
250
467
619
the initial pressure of ether was 312 mm Hg. Write
the rate equation for this reaction and determine the
rate constant of reaction.
Sol.
Q.45 From the following data show that decomposition of H2O2 in aqueous solution is first order.
Time (in minutes)
0
10
20
Volume (in c.c. of KMnO4) 22.8
13.3
8.25
Sol.
Q.46 A definite volume of H2O2 under going spontaneous decomposition required 22.8 cc. of standard
permanganate solution for titration. After 10 and 20
minutes respectively the volumes of permanganate
required were 13.8 and 8.25 c.c.
(a) Find order of reaction. How many the result be
explained ?
(b) Calculate the time required for the decomposition
to be half completed.
(c) Calculate the fraction of H2O2 decomposed after
25 minutes.
Sol.
Q.47 The following data were obtained in experiment on inversion of cane sugar.
Time (minutes) 0
60
120
180
360
Angle of
+13.1 +11.6 +10.2 +9.0 +5.87
–3.8
rotation (degree)
Show that the reaction is of first order. After what
time would you expect a zero reading in polarimeter ?
Sol.
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CHEMICAL KINETICS
Q.48 In the hydrolysis of propyl acetate in the presence of dilute hydrochloric acid in dilute aqueous solution the following data were recorded :
Time from start in minutes
60
350
Percentage of ester decomposed 18.17
69.12
Calculate the time in which half the ester was decomposed.
Sol.
Q.49 Hydrogen peroxide solution was stored in a mild
steel vessel. It was found, however, that the hydrogen peroxide decomposed on the walls of the vessel
(a first order reaction). An experiment with 100 ml of
a solution gave 10.31 ml oxygen (corrected to 1 atm
& 273 K) after 5.1 days under similar storage conditions. Find how long the peroxide can be stored before the loss of 20.00 ml oxygen occurs (per 100 ml
solution) if complete decomposition of the sample to
H2O2 gave 46.34 ml oxygen.
Sol.
Page # 137
PARALLEL AND SEQUENTIAL REACTION
y
k1
Q.51
For a reaction x
, calculate value of rak2
tio,
z
[ x] t
at any given instant t.
[ y] [ z]
Sol.
Q.52
late
k1
B
k2
C
A
k 1 = x hr–1 ; k1 : k2 = 1 : 10. Calcu-
[C]
after one hour from the start of the reaction.
[ A]
Assuming only A was present in the beginning.
Sol.
Q.53 A substance undergoes first order decomposition. The decomposition follows two parallel first orB
k1
der reactions as A
Q.50 The reaction given below, rate constant for
disappearance of A is 7.48 × 10–3 sec–1. Calculate the
time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to
0.145 atm and also find the total pressure after 100
sec.
2A (g)
4B(g) + C (g)
Sol.
; k 1 = 1.26 × 10–4 sec–1
k2
C
and k2 = 3.6 × 10–5 sec –1. Calculate the % distribution
of B & C.
Sol.
Q.54 For a reaction A
B
C t1/2 for A & B are 4
and 2 minutes respectively. How much time would be
required for the B to reach maximum concentration.
Sol.
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CHEMICAL KINETICS
TEMPERATURE DEPENDENCE OF RATE
Q.55
For the two parallel reactions A
k1
B and
(a) rate constant for the reaction at 27°C & 47° C and
(b) energy of activation for the reaction.
Sol.
A
C , show that the activation energy E' for the
disappearance of A is given in terms of activation
energies E1 and E2 for the two paths by
k2
E'
k1E1 k 2E2
k1 k 2
Sol.
Q.60 A catalyst lowers the activation energy for a
certain reaction from 75 kJ to 25 kJ mol–1. What will
be the effect on the rate of reaction at 25°C, after
things being equal.
Sol.
(ACTIVATION ENERGY)
Q.56 The energy of activation of a first order reaction is 104.5 kJ mole–1 and pre-exponential factor (A) is
5 × 10–13 sec–1. At what temperature, will the reaction
have a half life of 1 minute ?
Sol.
Q.57 The specific rate constant for a reaction increases by a factor of 4, if the temperature is changed
from 27°C to 47°C. Find the activation energy for the
reaction.
Sol.
Q.61 Given that the temperature coefficient for the
saponification of ethyl acetate by NaOH is 1.75. Calculate activation energy for the saponification of ethyl
acetate.
Sol.
Q.62 The rate constants of a reaction at 500 K and
700 K are 0.02 s–1 and 0.07 s–1 , respectively. Calculate the values of Ea and A at 500 K.
Sol.
Q.58 The energy of activation and specific rate constant for a first order reaction at 25°C are 100 kJ/
mole and 3.46 × 10–5 sec–1 respectively. Determine
the temperature at which half life of the reaction is 2
hours.
Sol.
MECHANISM OF REACTION
Q.63 The reaction 2NO + Br2 2NOBr, is supposed
to follow the following mechanism
Q.59 A first order reaction is 50% complete in 30
minutes at 27°C and in 10 minutes at 47°C. Calculate
the
(i) NO + Br2 fast
NOBr2
(ii) NOBr2 + NO slow 2NOBr
Suggest the rate of law expression.
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CHEMICAL KINETICS
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Sol.
Sol.
Q.64 For the reaction 2H2 + 2NO
N2 + 2H2O, the
following mechanism has been suggested :
2NO
N2O2 + H2
N 2O2 equilibrium constant K1 (fast)
K2
N 2O + H2O (slow)
N2O + H 2 K 3
N2 + H2O (fast)
Establish the rate law for given reaction.
Sol.
67. The reaction of formation of phosgene from CO
and Cl2 is CO + Cl 2 COCl 2 The proposed mechanism
is
k1
(i) Cl2
2 Cl (fast equilibrium)
k –1
k2
(ii) Cl + CO
COCl (fast equilibrium)
k –2
(iii) COCl + Cl 2 K 3
COCl2 + Cl (slow)
Show that the above mechanism leads to the followi ng
Q.65 Reaction between NO and O2 to form NO2 is
2NO + O2
2NO2 follows the following mechanism
K1
N2O2 (in rapid equilibrium)
NO + NO
K–1
K2
N2O2 + O2
2NO2 (slow)
Show that the rate of reaction is given by
1 d[NO2 ]
2
dt
K
rate
k3.
d[ COCl 2 ]
dt
law
k2
k1
k 2 k 1
K[ CO][ Cl 2 ] 3 / 2
Where
1/ 2
Sol.
K[NO] 2 [ O2 ]
Sol.
68. The approach to the following equilibrium was
observed kinetically from both directions :
PtCl2–
4 + H2O
[Pt (H2 O)Cl–3 ] + Cl– at 25°C, it was
found that
–
66. For the mechanism A + B
k1
k2
3
C;
C
k3
D
(a) Derive the rate law using the steady-state approximation to eliminate the concentration of C.
t
[PtCl 24 – ] = [3.9 × 10–5 sec–1] [PtCl24 – ] – [2.1 × 10–
L.mol–1 sec–1] × [Pt(H2 O)Cl 3 ] – [ Cl – ] What is the value
of equilibrium constant for the complexation of the
fourth Cl– by Pt(II) ?
Sol.
(b) Assuming that k3 < < k2, express the pre-exponential factor A and Ea for the apparent second-order
rate constant in terms of A1 , A2 and A3 and Eal, Ea 2
and Ea 3 for the three steps.
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CHEMICAL KINETICS
69. A solution of A is mixed with an equal volume of a
71. A certain reactant Bn+ is getting converted to
solution of B containing the same number of moles,
B(n+4)+ in solution. The rate constant of this reaction
and the reaction A + B = C occurs. At the end of 1h,
is measured by titrating a volume of the solution with
A is 75% reacted. How much of A will be left unreacted
a reducing reagent which only reacts with Bn+ and B(n
at the end of 2 h if the reaction is (a) first order in A
+ 4)+
and zero order in B; (b) first order in both A and B ;
B(n+4)+ to B(n–1)+. At t = 0, the volume of the reagent
and (c) zero order in both A and B ?
Sol.
consumed is 25 ml and at t = 10 min, the volume used
. In this process, it converts Bn+ to B(n–2)+ and
up is 32 ml. Calculate the rate constant of the conversion of Bn+ to B (n
4)
assuming it to be a first order
reaction.
Sol.
70. The decomposition of a compound P, at temperature T according to the equation
2P( g)
4Q( g ) R( g)
S( l) is the first order reaction. Af-
ter 30 minutres from the start of decomposition in a
72. A metal slowly forms an oxide film which com-
closed vessel, the total pressure developed is found
pletely protects the metal when the film thickness is
to be 317 mm Hg and after a long period of time the
3.956 thousand ths of an inch. If the film thickness is
total pressure observed to be 617 mm Hg. Calculate
1.281 thou. in 6 weeks, how much longer will it be
the total pressure of the vessel after 75 mintute, if volume of liquid S is supposed to be negligible. Also calculate
the time fraction t 7 / 8 .
before it is 2.481 thou? The rate of film formation
follows first order kinetics.
Sol.
Given : Vapour pressure of S(l) at temperature T =
32.5 mm Hg
Sol.
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CHEMICAL KINETICS
Page # 141
73. A vessel contains dimethyl ether at a pressure of
75. The gaseous reaction : n1A(g)
0. 4
as
order with respect to A. It is studied at a constant
CH4 ( g) CO(g) H2 (g) . The rate constant
pressure, with a0 as the initial amount of A.Show that
at m.
CH3 OCH3 ( g)
Dimeth yl
ethe r
deco mp oses
of decomposition is 4.78 × 10–3 min–1. Calculate the
ratio of initial rate of diffusion to rate of diffusion
after 4.5 hours of initiation of decomposition. Assume
the composition of gas present and composition of
n2B(g) is first
the volume of system at the concentration of A at
time 't' are given by the expressions.
V
V0
n2
n
– 2 – 1 exp(–n1kt )
n1
n1
gas diffusing to be same.
Sol.
[ A] t
[ A] 0
;
exp(–n1kt )
n2
–
n1
n2
– 1 exp(–n1kt )
n1
Sol.
74. For the following first order gaseous reaction
k1
2B(g)
A(g)
k2
C(g)
The initial pressure in a container of capacity Vlitres
is 1 atm. Pressure at time t = 10 sec is 1.4 atm and
after infinite time it becomes 1.5 atmosphere. Find
the rate constant k1 and k2 for the appropriate reactions.
Sol.
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Page # 142
CHEMICAL KINETICS
EXERCISE – IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
Q.1 For the reaction H 2 +
relationship is -
2
2H the true
[AIEEE-2002]
d[ H 2 ]
d[ 2 ]
d[ H ]
=
=
dt
dt
dt
d [H 2 ]
d[ 2 ]
1 d [H ]
(B)
=–
=
dt
dt
dt
2
2d [ H 2 ]
2d[ 2 ]
d [H ]
(C) –
=–
=
dt
dt
dt
2d[ H2 ]
2d[ 2 ]
1 d[ H ]
(D)
=–
=
dt
dt
dt
2
(A) –
Q.4 In the equation Kt = log C0 – log Ct, the curve
between t and log Ct is [AIEEE-2002]
(A) a straight line
(B) a parabola
(C) a hyperbola
(D) none
Sol.
Sol.
Q.2 A chemical reaction
[2A] + [2B] + [C]
product follows the rate equation : r [A] [B]2 then
order of reaction is [AIEEE-2002]
(A) 0
(B) 1
(C) 2
(D) 3
Sol.
Q.3 The unit of rate constant of first & second order
reaction is respectively [AIEEE-2002]
(A) time–1 , mole–1 . litre . time–1
(B) mole ltr–1, time –1
(C) mole–1 . litre. time–1 , time–1
(D) sec–1 , litre–1
Sol.
Q.5 Consider following two reactions
d[ A]
A
Product
–
= k1 [A]º
dt
d [B]
B
Product
–
= k2 [B]
dt
Units of k1 and k2 are expressed in terms of molarity (mol
L–1) and time (sec–1) as –
[AIEEE-2002]
(A) sec–1 , M sec–1
(B) M sec–1 ,M sec–1
(C) sec–1 , M–1 sec–1 (D) M sec–1 , sec–1
Sol.
Q.6 H2 gas is adsorbed on the metal surface like
tungsten. This follows............ order reaction –
[AIEEE-2002]
(A) Third
(B) Second
(C) Zero
(D) First
Sol.
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CHEMICAL KINETICS
Page # 143
Q.7 In respect of the equation k = Ae –Ea/RT in chemical
kinetics, which one of the following statement is correct?
[AIEEE-2003]
(A) Ea is energy of activation
(B) R is Rydberg’s constant
(C) k is equilibrium constant
(D) A is adsorption factor
Sol.
Q.8 For the reaction system:
2NO(g) + O2(g)
2NO2 (g) volume is suddenly reduced to half its value
by increasing the pressure on it. If the reaction is of
first order with respect to O2 and second order with
repect to NO, the rate of reaction will –[AIEEE-2003]
(A) Increase to eight times of its initial value
(B) Increase to four times of its initial value
(C) Decrease to one-fourth of its initial value
(D) Decrease to one-eighth of its initial value
Sol.
Q.9 The rate law for a reaction between substances
A and B is given by
Rate = k [A]n [B]m On doubling the concentration of
A and halving the concentration of B, the ratio of the
new rate to the earlier rate of the reaction will be as–
[AIEEE-2003]
Q.10 In a first order reaction, the concentration of
the reactant, decreases from 0.8 M to 0.4 M in 15
minutes. The time taken for the concentration to
change from 0.1 M to 0.025 M is [AIEEE-2004]
(A) 30 minutes
(B) 15 minutes
(C) 7.5 minutes
(D) 60 minutes
Sol.
Q.11 The rate equation for the reaction 2 A + B C
is found to be : rate = k [A] [B] . The correct
statement in relation to this reaction is that the
[AIEEE-2004]
(A) unit of k must be s–1
(B) t1/2 is a constant
(C) rate of formation of C is twice the rate of
disappearance of A
(D) valu e of k is i nd ep end ent of th e init ial
concentrations of A and B
Sol.
,
(A) (n–m)
Sol.
(B) 2(n–m)
(C)
1
2
(m n)
(D) (m+n)
Q.12 Consider an endothermic reaction X
Y with
the activation energies Eb and Ef for the backard and
forward reactions, respectively. In general
[AIEEE-2005]
(A) Eb > Ef
(B) Eb < Ef
(C) there is no definite relation between Eb and Ef
(D) Eb = Ef
Sol.
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CHEMICAL KINETICS
Q.13 A reaction involing two different reactants can
never be –
[AIEEE-2005]
(A) first order reaction
(B) unimolecular reaction
(C) bimolecular reaction
(D) second order reaction
Sol.
Q.16 Rate of reaction can be expressed by Arrhenius
equation as k = Ae–E/RT , In this equation, E represents
[AIEEE 2006]
(A) the energy below which colliding molecules will
not react
(B) the total energy of the reacting molecule at a
temperature, T
(C) the fraction of molecules with energy greater than
the activation energy of the reaction
(D) the energy above which all the colliding molecules
will react
Sol.
Q.14
Q.17 The following mechanism has been proposed
for the reaction of NO with Br2 to form NOBr :
NO(g) + Br2 (g)
NOBr2 (g)
NOBr2 (g) + NO (g) 2 NOBr (g)
If the second step is the rate determining step, the
order of the reaction with respect to NO (g) is [AIEEE 2006]
(A) 0
(B) 3
(C) 2
(D) 1
Sol.
t 1 can be taken as the time taken for the
4
3
concentration of a reactant to drop to
of its initial
4
value. If the rate constant for a first order reaction is
K, t 1 can be written as –
[AIEEE-2005]
4
(A) 0.29/K
(C) 0.75/K
Sol.
(B) 0.10/K
(D) 0.69/K
Q.15 A reaction was found to be second order with
respect to the concentration of carbon monoxide. If
the concentration of carbon monoxide is doubled, with
everything else kept the same, the rate of reaction
will [AIEEE 2006]
(A) triple
(B) increase by a factor of 4
(C) double
(D) remain unchanged
Sol.
Q.18 The energies of activation for forward and
reverse reactions for A2 + B2
2AB are 180 kJ
mol–1 and 200 kJ mol –1 respectively. The presence
of a catalyst lowers the activation energy of both
(forward and reverse) reactions by 100 kJ mol–1 .
The enthalpy change of the reaction (A2 + B2
2A B) i n the presence of catalyst will be
(in kJ mol–1) –
[AIEEE 2007]
(A) 300
(B) 120
(C) 280
(D) 20
Sol.
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CHEMICAL KINETICS
Page # 145
Q.19 Consider the reaction, 2A + B
Products.
When concentration of B alone was doubled, the
half -lif e of B d id n ot c hang e. W hen the
concentration of A alone was doubled, the rate
increased by two times. The unit of rate constant
for this reaction is –
[AIEEE 2007]
(A) L mol –1 s–1
(B) no unit
(C) mol L–1 s–1
(D) s–1
Sol.
Q.22 The time for half life period of a certain reaction
A products is 1 hour. When the initial concentration
of the reactant ‘A’, is 2.0 mol L–1, how much time does
it take for its concentration to come from 0.50 to
0.25 mol L–1 if it is a zero order reaction ?
(A) 4 h (B) 0.5 h (C) 0.25 h (D) 1 h
Sol.
[AIEEE 2010]
1
A 2B, rate of disappearance
2
of ‘A’ related to the rate of appearance of ‘B’ by the
expression [AIEEE 2008]
Q.20
(A)
(C)
For a reaction
d[A]
dt
d[A]
dt
1 d[B]
4 dt
d[B]
4
dt
(B)
(D)
d[A]
dt
d[A]
dt
d[B]
dt
1 d[B]
2 dt
Sol.
Q.23 The rate of a chemical reaction doubles for
every 10ºC rise of temperature. If the temperature
is raised
by 50ºC, the rate of the reaction increases by about
:
[AIEEE 2011]
(A) 24 times (B) 32 times (C) 64 times (D) 10 times
Sol.
Q.21 The half life period of a first order chemical
reaction is 6.93 minutes. The time required for the
completion of 99% of the chemical reaction will be
(log 2 = 0.301)
[AIEEE 2009]
(A) 23.03 minutes (B) 46.06 minutes
(C) 460.6 minutes (D) 230.3 minutes
Sol.
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Page # 146
Q.24 For a fi rs t orde r (A )
Prod uc ts , the
concentration of A changes from 0.1 M to 0.025
M in 40 minutes. The rate of reaction when the
concentration of A is 0.01 M, is :
[AIEEE 2012]
(A) 3.47 × 10–5 M/min
(B) 1.73 × 10–4 M/min
(C) 1.73 × 10–5 M/min
(D) 3.47 × 10–4 M/min
Sol.
Q.25 The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation
energy of such a reaction will be:
(R = 8.314 JK–1 mol –1 and log 2 = 0.301)
(A) 58.5 kJ mol –1
[IIT Mains 2013]
(B) 60.5 kJ mol –1
(C) 53.6 kJ mol –1
(D) 48.6 kJ mol –1
Sol.
CHEMICAL KINETICS
Q.26
For the non-stoichiometre reaction 2A + B
C + D, the following kinetic data were obtained
in three separate expreiments, all at 298 K
Initial
Initial
Initial rate of
Concentration Concentration formation of C
(A)
(B)
(mol L S )
0.1 M
0.1 M
1.2 10 3
0.1 M
0.2 M
1.2 10 3
0.2 M
0.1 M
2.4 10 3
the rate law for the formation of C is
[AIEEE-2014]
(A)
dc
= k[A] [B]2
dt
(B)
dc
= k[A]
dt
(C)
dc
= k[A] [B]
dt
(D)
dc
= k[A]2 [B]
dt
Sol.
Q.27
Higher order (>3) reactions are rare due to:
(A) shifting of equilibrium towards reactant due
to elastic collisions
(B) loss of active species on collision
(C) low probability of simultaneous collision of
all the reacting species
(D) increas e in entropy and ac tivation
energy as more molecules are involved
[AIEEE - 2015]
Sol.
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CHEMICAL KINETICS
LEVEL – II
OBJECTIVE
1. The rate constant for the reaction
2N2O5 4NO2 + O2
is 3.0 × 10–5 sec–1 . If the rate is 2.4 × 10–5 mol litre–1
sec–1, then the concentration of N2O5 (in mol litre–1) is
[JEE SCR 2000]
(A) 1.4
(B) 1.2
(C) 0.004
(D) 0.8
Sol.
2. If I is the intensity of absorbed light and C is the
concentration of AB for the photochemical process
AB + hv
AB*, the rate of formation of AB* is
directly proportional to
[JEE SCR 2001]
(A) C
(B) I
(C) I2
(D) Cl
Sol.
3. Consider the chemical reaction, N2(g) + 3H2(g)
2NH3(g). The rate of this reaction can be expressed in
term of time derivative of concentration of N2(g), H2(g)
or NH3(g). Identify the correct relationship amongst
the rate expressions.
[JEE SCR 2002]
(A) Rate = –
d[N2 ]
1 d[H 2 ]
1 d[NH3 ]
=–
=
3 dt
dt
2 dt
(B) Rate = –
3 d [H 2 ]
2 d[NH 3 ]
d[N2 ]
=–
=
dt
dt
dt
(C) Rate =
(D) Rate =
Page # 147
JEE ADVANCED
Sol.
4. In a first order reaction the concentration of
reactant decreases from 800 mol/dm3 to 50 mol/dm3
in 2 × 104 sec. The rate constant of reaction in sec–1
is :
[JEE SCR 2003]
4
(A) 2 × 10
(B) 3.45 × 10–5
–4
(C) 1.3486 × 10
(D) 2 × 10–4
Sol.
5. The reaction, X
Product follows first order
kinetics. In 40 minutes the concentration of X changes
from 0.1 M to 0.025 M. Then the rate of reaction
when concentration of X is 0.01 M [JEE SCR 2004]
(A) 1.73 × 10–4 M min–1
(B) 3.47 × 10–5 M min–1
(C) 3.47 × 10–4 M min–1
(D) 1.73 × 10–5 M min–1
Sol.
1 d[H2 ]
1 d[NH3 ]
d[N2 ]
=
=
dt
3 dt
2 dt
d[N 2 ]
– d[H2 ]
d[NH3 ]
=
=
dt
dt
dt
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Page # 148
6. Which of the following statement is incorrect about
order of reaction ?
[JEE 2005]
(A) Order of reaction is determined experimentally
(B) It is the sum of power of concentration terms in
the rate law expression
(C) It does not necessarily depend on stoichiometric
coefficients.
(D) Order of the reaction can not have fractional value.
Sol.
7. Consider a reaction aG + bH
Products. When
concentration of both the reactants G and H is
doubled, the rate increases by eight times. However,
when concentration of G is doubled keeping the
concentration of H fixed, the rate is doubled. The
overall order of the reaction is
[JEE 2006]
(A) 0
(B) 1
(C) 2
(D) 3
Sol.
CHEMICAL KINETICS
8. Under the same reaction conditions, initial
concentration of 1.386 mol dm–3 of a substance
becomes half in 40 seconds and 20 seconds through
first order and zero order kinetics, respectively. Ratio
k1
k 0 of the rate constants for first order (k 1) and
zero order (k0) of the reactions is
[JEE 2008]
(A) 0.5 mol–1 dm3
(B) 1.0 mol dm–3
(C) 1.5 mol dm–3
(D) 2.0 mol–1 dm3
Sol.
9. For a first order reaction A
P, the temperature
(T) dependent rate constant (k) was found to follow
the equation log k = – (2000) 1/T + 6.0. The preexponential factor A and the activation energy Ea,
respectively, are
(A) 1.0 × 106s–1 and 9.2 kJ mol–1
(B) 6.0 s–1 and 16.6 kJ mol–1
(C) 1.0 × 106 s–1 and 16.6 kJ mol –1
(D) 1.0 × 106 s–1 and 38.3 kJ mol –1
[JEE 2009]
Sol.
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CHEMICAL KINETICS
Page # 149
10. Plots showing the variation of the rate constant
(k) with temperature (T) are given below. The plot
that follows Arrhenius equation is
[JEE 2010]
T
k
(A)
(B)
k
T
k
SUBJECTIVE PROBLEM
12. A hydrogenation reaction is carried out at 500 K.
If the same reaction is carried out in the presence of
a catalyst at the same rate, the temperature required
is 400 K. Calculate the activation energy of the reaction
if the catalyst lowers the activation barrier by 20 kJ
mol–1.
[JEE 2000]
Sol.
k
(C)
(D)
T
T
Sol.
11. For the first order reaction
[JEE 2011]
2N2O5(g) 4NO2 (g) + O2(g)
(A) The concentration of the reactant decreases exponentially with time.
(B) The half-life of the reaction decreases with increasing temperature.
(C) The half-life of the reaction depends on the initial
concentration of the reactant.
(D) the reaction proceeds to 99.6% completion in
eight half-life duration.
Sol.
13. The rate of first order reaction is 0.04 mole litre–1 s–1
at 10 minutes and 0.03 mol litre–1 s–1 at 20 minutes
after initiation. Find the half life of the reaction.
[JEE 2001]
Sol.
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Page # 150
CHEMICAL KINETICS
14.
2X(g) 3Y(g) + 2Z(g)
Time (in Min)
0
100
200
Partial pressure of X
800
400
200
(in mm of Hg)
Assuming ideal gas condition. Calculate
(A) Order of reaction
(B) Rate constant
(C) Time taken for 75% completion of reaction
(D) Total pressure when Px = 700 mm.
[JEE 2005]
Sol.
16. An organic compound undergoes first-order
decomposition. The time taken for its decomposition
to 1/8 and 1/10 of its initial concentration are t1/8
and t1/10 respectively.
What is the value of
15. The concentration of R in the reaction R P was
measured as a function of time and the following data
is obtained:
1. 0
0.75
0.4 0
0.1 0
t(min.)
0. 0
0.05
0.12
0.18
[JEE 2010]
The order of the reaction is
Sol.
[t1 / 10 ]
10 ? (take log10 2 = 0.3)
Sol.
17.
[R] (molar)
[t1 / 8 ]
[JEE 2012]
In the reaction,
P+Q
R+S
the time taken for
75% reaction of P is
twice the time taken
for 50% reaction of [Q]o
P. The concentration
of Q va ries w ith [Q]
re ac tion t ime as
shown in figure. the
overall order
of the reaction is
(A) 2
(B) 3
(C) 0
[JEE 2013]
Time
(D) 1
Sol.
Corporate
CorporateHead
HeadOffice
Office::Motion
MotionEducation
EducationPvt.
Pvt.Ltd.,
Ltd.,394
394--Rajeev
RajeevGandhi
GandhiNagar,
Nagar,Kota-5
Kota-5(Raj.)
(Raj.)
CHEMICAL KINETICS
Page # 151
19.
For the elementary reaction M
N, the rate
of disappearance of M increases by a factor of
8 upon doubling the concentration of M. The
order of the reaction with respect to M is :
[IIT - 2014]
(A) 4
(B) 3
(C) 2
(D) 1
Sol.
18.
The unbalanced chemical reactions given in list
show missing reagent or condition (?) which
are provided in List II. Match List I with List II
and select the correct answere using the code
given below the lists :
List I
List II
P. PbO2+H2SO4
?
Q. Na2 S2 O3 +H2O
R. N2 H4
S. XeF2
?
PbSO4+O2 other product 1. NO
?
NaHSO4+ other product 2. I2
N2 + other product
?
3. Warm
Xe + other product
4. Cl2
Codes :
(A)
(B)
(C)
(D)
4
3
1
3
P
2
2
4
4
Q
3
1
2
2
R
1
4
3
1
S
Sol.
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CHEMICAL KINETICS
Page # 96
Answer-Key
Exercise-I
1.
B
2.
D
3.
B
4.
D
5.
A
6.
B
7.
C
8.
D
9.
B
10.
D
11.
B
12.
C
13.
D
14.
A
15.
B
16.
D
17.
A
18.
A
19.
B
20.
B
21.
B
22.
B
23.
B
24.
C
25.
C
26.
C
27.
C
28.
B
29.
D
30.
D
31.
D
32.
D
33.
A
34.
B
35.
C
36.
D
37.
C
38.
B
39.
A
40.
C
41.
C
42.
C
43.
D
44.
A
45.
B
46.
B
47.
B
48.
C
49.
D
50.
A
Exercise-II
1.
A
2.
A
3.
A
4.
C
5.
C
6.
D
7.
C
8.
C
9.
A
10.
A,D
11.
A,B,D
12.
A,D
13.
C
14.
C
15.
D
16.
A
17.
C
18.
A
19.
B
20.
A
21.
B
22.
C,D
23.
A,B
24.
A,B
25.
A,C
26.
A,B,C,D
27.
A,B,C,D
28.
A,B,C
29.
B,D
30.
A,C
31.
B,C
32.
A,B,D
33.
B,C
34.
A,C,D
35.
A,B
36.
B
37.
A
38.
D
39.
C
40.
A
41.
B
42.
C
43.
A
44.
D
45.
A
46.
B
47.
A
48.
D
49.
A
50.
A–S, B–QST, C–QRS, D–QRS
51.
A
R, B
Q,
53.
A
P, B
Q,S
59.
3
C
C
P, D
S
R,T D
52.
A
S, B
R,
C
P, D
Q
V
INTEGER TYPE
54.
2
55.
20
56.
5
57.
10
58.
6
60.
3
61.
5
62.
6
63.
2
64.
6
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Page # 97
CHEMICAL KINETICS
Exercise-III
1.
(a) 0.04 M/sec
3.
2k1 = k 2 = 4k3
5.
rate increase by 27 times
6.
(i) r =
7.
(i) 7.2 mol litre–1 min–1, (ii) 7.2 mol litre–1 min–1 8.
9.
(i) 7.2 M, (ii) 10 M
13.
1.2 hr
1 d[NO]
4 dt
(b) 0.02
2.
(a) 0.019 mol L–1 s–1, (b) 0.037 mol L–1 s–1
4.
(i)
dx
= k[A] [B]2, (ii) rate increases by 8 times
dt
9 10 –4 mol litre –1 sec–1, (ii) 36 × 10–4 mol litre –1 sec–1, (iii) 54 × 10–4 mol litre –1 sec–1
10. K = 0.01 M min–1
14. (i) 36 min, (ii) 108 min.
1
6
11. 0.75 M
15. (i) 0.0223 min–1, (ii) 62.17 min
18. expiry time = 41 months 19. 3.3 × 10–4 s–1
20. 62.13 min
12.
6 × 10–9 sec
17. 924.362 sec.
21. 11.2 %
22.
(a) Third order, (b) r = k [NO]2 [H2 ], (c) 8.85 × 10–3 M sec –1
23.
(a) order w.r.t NO = 2 and w.r.t Cl2 = 1, (b) r = K [NO]2 [Cl 2], (c) K = 8 L2 mol –2 s–1
(d) rate = 0.256 mole L–1 s1
24.
(i) first order (ii) k = 1.308 × 10–2 min–1 (iii) 73%
25.
(i) rate = [A][B] ; (ii) k = 4 × 10–2 M–1 s–1 ; (iii) rate = 2.8 × 10–3 M s–1
26.
(i) Zero order, (ii) K =
28. (a) n = 1, (b)
5Pa
s
27.
Zero order
dx
= k[CH3COCH3], (c) 8.67 × 10–3 s–1, (d) 1.56 × 10–5 M s–1
dt
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CHEMICAL KINETICS
Page # 98
29.
166.6 min
30. 5.26%, 4.62 × 105 sec 32. (i) t = 13.96 hrs, (ii) 2.2176 litre
33.
54 min
34. 15 min
35.
K
P3
1
ln
K
t 2(P3 – P2 ) 36.
P3
1
ln
t (P3 – P2 )
37. K
38.
K
1
4V3
ln
K
t 5( V3 – V2 ) 39.
1
r
ln
t (r – r1 )
40. 8.12 × 10–6 Ms–1, 0.012 atm min–1
41.
(a) 90 mm, (b) 47 mm, (c) 6.49 × 10–2 per minutes, (d) 10.677 min.
42.
First order
44.
(i) r = K[(CH3)2 O], 0.000428 sec–1
46.
(a) first order, (b) 13.75 minutes, (c) 0.716
47.
966 min
51.
(K 1 K 2 )
43. k1 = 2.605 × 10–3 min–1
45. First order
48. 206.9 min
1
e
V1
1
ln
t (2V1 – V2 )
52.
t –1
56. 349.1 k
[C]
[ A]
10 11x
(e – 1)
11
57. 55.33 kJ mole–1
49. 11.45 days
53. 77.7, 22.3
54. t = 4 min
58. 306 k
59.
(a) 2.31 × 10–12 min–1, 6.93 × 10–2 min–1, (b) 43.85 kJ mole–1
60.
rate of reaction increases 5.81 × 108 times
63.
r = K [NO]2 [Br2]
66.
(a)
68. 53.84
d(D)
dt
50. 0.180 atm, 47.69 sec.
61.
10.272 k cal mol –1 62. 1.585
64. r = K [NO]2 [H2], where K = k2 × K 1
k1k 3 ( A )(B)
A1A 3
+
k 2 k 3 ; (b) Ea = Ea 1 Ea 3 – Ea2.A = A 2
69.
(a) 6.25 ; (b) 14.3 ;
71. 0.0207 min–1 72. 15.13 week
(c) 0%
73. 0.26 : 1
70. Pt = 379.55 mm Hg, t7/8 = 399.96 min
74.
0.0805
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Page # 99
CHEMICAL KINETICS
Exercise-IV
Level -1
1.
C
2.
D
3.
A
4.
A
5.
D
6.
C
7.
A
8.
A
9.
B
10.
A
11.
D
12.
B
13.
B
14.
A
15.
B
16.
A
17.
C
18.
D
19.
A
20.
A
21.
B
22.
C
23.
B
24.
D
25.
C
26.
B
27.
C
11.
ABD
16.
0009
Level -2
OBJECTIVE PROBLEM
1.
D
2.
B
3.
A
4.
C
5.
C
6.
D
7.
D
8.
A
9.
D
10.
A
SUBJECTIVE PROBLEM
12. 100 kJmol–1
13. t1/2 = 24.14 min
14. (B) 1 (B) 6.93 × 10–3 min (C) 200 min (D) 950 mm
17.
D
18.
D
19.
15. 0
B
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FUNCTION
Page # 156
FUNCTION
A.
DEFINITION
Function is defined as a rule or a manner or a mapping or a correspondence f which maps each & every
element of set A with a unique element of set B. It is denoted by :
f:A
B
or A
f
B we read it as “ f is a function from A to B”
f
Ex.
f maps A to B
f
A
B
1
2
3
4
w
x
y
z
A
B
w
x
y
z
1
Yes
2
Figure (1)
No
Figure (2)
f
Figure –2 does not represent a function because conversion
is allowed (figure–3) But diversion is not allowed.
A
B
1
2
3
4
w
x
Yes
Figure (3)
Ex.1
Which of the following correspondences can be called a function ?
(A) f(x) = x3
(C) f(x) =
; {–1, 0, 1}
x ; {0, 1, 4}
{0, 1, 2, 3}
{–2, –1, 0, 1, 2}
(B) f(x) =± x ; {0, 1, 2}
{–2, –1, 0, 1, 2}
(D) f(x) = – x ; {0, 1, 4}
{–2, –1, 0, 1, 2}
Sol.
f(x) in (C) & (D) are functions as definition of function is satisfied. while in case of (A) the given
relation is not a function, as f(–1) codomain. Hence definition of function is not satisfied. While in
case of (B), the given relation is not a function, as f(1) = ± 1 and f(4) = ± 2 i.e. element 1 as well as
4 in domain are related with two elements of codomain. Hence definition of function is not satisfied.
Ex.2
If X = {a, b, c, d, e} & Y = {p, q, r, s, t} then which of the following subset(s) of X × Y is/are a
function from X to Y.
(A) {(a, r) (b, r) (b, s) (d, t) (e, q) (c, q)}
(B) {(a, r) (b, p) (c, t) (d, q)}
(C) {(a, p) (b, t) (c, r) (d, s) (e, q)}
(D) {(a, r) (b, r) (c, r) (d, r) (e, r)}
Let us check every option for the two conditions of the function
(A)
b has two output (images) namely r & s
Not a function
(B)
e X does not have any image
Not a function
(C)
every element of X has one and only one output
it is a function
(D)
every element’s output is r
it is a function
Hence correct options are (C) & (D).
Sol.
“Function” as an ordered pair :
f:A
B
f : {(1, x), (2, x), (3, x)}
f A×B
where A × B is the cartesian product of two set A & B
= {(1, x), (2, x), (3, x), (1, y), (2, y), (3, y)}
A
1
2
3
f
B
x
y
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FUNCTION
Page # 157
Remark : Every function from A to B satisfied the following Relation :
(1)
f A×B
(2)
Notation :
a
A
(For all)
b
B | b = f(a)
(there exist) (such that)
(3)
(a, b) f & (a, c) f
b=c
(4)
In a graphical representation of a function y = f(x).
If vertical line cuts the curve more than once then it is not a function. It is called as vertical line test
y2 = 4ax
y = f(x)
Ex :
It is not a function as vertical line touches
curve at more than once point.
B.
DOMAIN, CO-DOMAIN & RANGE OF A FUNCTION
Let f : A
B, then the set A is known as the domain of f& the set B is known as co-domain of f. If a
member ‘a’ of A is associated to the member ‘b’ of B, then’b’ is called the f-image of ‘a’ and we write
b = f(a). Further ‘a’ is called a pre-image of ‘b’. The set {f(a):
a A} is called the range of f and
is denoted by f(A). Clearly f(A) B.
If only expression of f(x) is given (domain and codomain are not mentioned), then domain is set of those
values of ‘x’ for which f (x) is real, while codomain is considered to be (–
) (except in ITFs)
A function whose domain and range are both subsets of real numbers is called a real function.
(Algebraic Operations on Functions) : If f & g are real valued functions of x with domains A and B
respectively, then both f & g are defined in A B . Now we define f + g , f g , (f . g) & (f/g) as follows:
(i)
(f ± g) (x) = f(x) ± g(x)
(ii)
(iii)
(f g) (x) = f(x) g(x)
domain in each case is A
f
f( x)
(x) =
domain is {x x
g
g( x )
Ex.3
Find the domain of definition of the function
Sol.
For y to be defined
(i)
x 5
x
2
A
B
B and g(x)
y = log10
0} .
x 5
x 2 10 x 24
3
x 5
0
10x 24
When x – 5, x = 5 and when x2 – 10 x + 24, x = 4, 6
sign scheme for
Put x = 0
x
2
x 5
is as follows.
10 x 24
x 5
x
2
0
10x 24
–
–ve
4
+ve
4 < x < 5 or x > 6
5
–ve
6
+ve
...(A)
1
(ii)
( x 5) 3 is defined for all x
Combining (A) and (B), we get 6 < x <
...(B)
or 4 < x < 5
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[
FUNCTION
Page # 158
Ex.4
Find the domain of the function f(x) =
Sol.
x2 – x – 6
0
and
(x – 3) (x + 2)
x
3 or x
x
(– , –2]
x
6–x
0
–2
[3,
(– , –2]
x2
6 x
0
x
6
x
(– , 6]
–2
)
Find the domain of given function f(x) =
Sol.
3x – x3
x3 – 3x
C.
(
6
0
x(x2 – 3)
3x x3
0
x(x –
3 ) (x +
–
x
–3
[3, 6]
Ex.5
0
x 6 +
,
3]
[0,
3]
3)
0
–
+
3
+
3
IMPORTANT TYPE OF FUNCTIONS
(1) Trigonometric function :
Function
Domain
Range
Curve
y = sin x
1
(i) f(x) = sin x
x
R
y
3 /2
/2
[–1, 1]
2
3 /2
x
2
/2
–1
y = cos x
1
3 /2
/2
2
(ii) f(x) = cos x
x
R
y
3 /2
2
/2
[–1, 1]
x
5 /2
–1
y = tan x
3
2
(iii) f(x) = tan x
x
R – (2n + 1)
2
,n
y
O
2
3
2
2
R
x
y = cot x
(iv) f(x) = cot x
x
R–n
n
y
3
2
R
O
2
2
2
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x
FUNCTION
Page # 159
y = cosec x
1
(v) f(x) = cosec x
x
R –n , n
y
, –1]
[1,
0
/2
)
/2
3 /2
x
2
–1
y = sec x
1
(vi) f(x) = sec x
(2)
(3)
x R – (2n+1)
2
,n
y
, –1]
[1,
)
/2
3 /2
0
/2
x
3 /2
–1
Polynomial Function :
f(x) = a0xn + a 1xn – 1 + a 2 xn–2 +.......+ an
where a 0, a1, a2 ........ a n R
n W
If a 0 0, then f(x) is called nth degree polynomial and Domain x
R
Algebraic Function : A function is called an algebraic function. If it can be constructed using
algebraic operations such as additions, subtractions, multiplication, division taking roots etc.
All polynomial functions are algebraic but converse is not true.
Ex :
f(x) =
x4
3
3/5
3
2
5x2 + x + (x + 5) , f(x) = x + 3x + x + 5
Remark : Function which are not algebraic are called as TRANSCENDENTAL FUNCTION.
(4)
Ex :
f(x) =
Ex :
f(x) =
5x2 ) 3 / 5
x3
x2
+ 3 x2
5x 6 + n x
x 7
7 + e nx +
x2 7
0
Ex.
f(x) =
x 4 3x2
x
2
transcidental function
algebraic function.
Rational Function : It is a function of form f(x) =
and h(x)
(5)
( x5
g( x)
, where g(x) & h(x) are poly. function
h( x)
2
4
Logarithmic function : f(x) = logax, where x > 0, a > 0, a 1
a
base, x number or argument of log.
Case–I : 0 < a < 1
Case–II : a > 1
f(x)
f(x) = loga x
Domain : x (0,
Range : y R
f(x)
f(x) = n x
(1, 0)
)
O
x
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O
(1, 0)
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x
FUNCTION
Page # 160
(6)
Exp on ent ial f un ct ion : f(x) = ax, where a > 0, a 1
a
Base
x
Exponent
Case–I : 0 < a < 1 ; a = 1/2
1
2
f(x) =
(7)
(1, 0)
R
Range : y
(0,
(0,1)
O
x
x
)
Absolute value function (Modulus function) :
y
x ; x 0
x ; x 0
Domain : x
y= x
y=–x
R; Range : y
R+
x
{0}
Signum function :
y
1
1; x 0
0 ; x 0
1; x 0
y = sgn (x) =
Domain : x
(9)
f(x)
f(x)
x
Domain : x
y = |x| =
(8)
Case–II : a > 1
R; Range : y
0
x
–1
{–1, 0, 1}
Greatest integer function (step-up function) :
y
y = f(x) = [x]
2
x
; x I
Greatest Integer ; otherwise
less than x
1
0
Domain : x R; Range : y I
Ex :
[2 . 3] = 2, [5] = 5, [–2 . 3] = –3
1
2
3
x
–1
Properties :
(i)
[x]
x < [x] + 1
(ii)
(10)
Fractional part function :
[x + m] = [x] + m ; m
I
(iii)
y = f(x) = {x} = x – [x]
y
Domain : x
1
R; Range : [0, 1)
Ex : 2.3 = 2 + 0.3
0 ; x I
1; x I
[x] + [–x] =
fractional part
–1
–2
0
1
2
3
x
Integer part
Properties :
(i)
Fractional part of any integer is zero.
(iii)
0; x I
{x} + {–x} = 1 ; otherwise
(ii)
{x + n} = {x}, n
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FUNCTION
Page # 161
Ex.6
Find the range of the following functions : (a) y =
Sol.
(a)
We have
since, sin 3 x
i.e.
i.e.
(b)
1
2
2
gives
Ex.7
Sol.
1–
6
1
2
We have
i.e.
2
y
sin 3x
i.e.
1
2
y
1, therefore we have
4
Now, we have 2
i.e.
1
–2
y
1 –2
y
– 2
1
1
i.e. sin 3x + cos 3x =
y –2
2 sin 3x cos 3x
y=
2 sin 3x 4
i.e.
x2 1
x2 2
1
(b) y = sin–1
2 sin 3x cos 3x
2–
2
2
1
y
2
Hence, the range is y
2
1
2
y
1
2
4
2
1
2
1
,
2 2
2
.
x2 1
1
1 2
2
x 2
x 2
x2 + 2 <
1
2
i.e.
1
x2
2
x2 1
<1
x2 2
i.e. sin–1
i.e.
y
Hence, the range is
2
1
i.e.
1
1
1 2
<1
2
x
2
1
2
1
2
>0
y
x2
2
1
2
<0
sin–1
x2 1
< sin–1 1
x2 2
, .
6 2
Find the range of following functions : (i) y = ln (2x – x2) (ii) y = sec–1 (x 2 + 3x + 1)
(i)
using maxima–minima, we have
(2x – x2) (– , 1]
For log to be defined accepted values are 2x – x2 (0, 1]
{i.e. domain (0, 1]}
ln (2x – x2) (0, 1]
range is (– , 0]
(ii)
y = sec–1 (x2 + 3x + 1)
y
Let t = x2 + 3x + 1 for x
R then
but y = sec–1 (t)
t
from graph range is y
0,
5
,
4
t
sec–1 (–5/4)
5
, 1
4
2
sec
[1,
1
/2
)
5
,
4
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–1
t
0
1
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FUNCTION
Page # 162
Ex.8
Find the range of y = ln(sin
Sol.
We have
1
x2 + x + 1 = x
x2
1
2
Also, for the function y = ln(sin
Thus, we have
Sol.
x2 + x + 1
2
+
3
which is a positive quantity whose minimum value is 3/4.
4
x2
x 1) to be defined, we have x2 + x+ 1 1
3
2
1 i.e.
x2
x 1
i.e.
3
sin 1 ( x2
x 1)
2
sin–1 x is an increasing function, the inequality sign remains same]
[
Ex.9
3
4
1
x 1)
i.e.
ln
i.e.
0.046
f:R
R, f (x) =
3
ln(sin–1 x 2
x 1
ln(sin–1
x 1 )] = 0.
3x 2
x2
ln
2
Hence, the range is y
[ n /3, n /2]
mx n
. If the range of this function is [– 4, 3) then find the value of (m2 + n2).
x 1
2
mx n 3
3( x2 1) mx n 3
f (x) =
;
f
(x)
=
3
+
1 x2
1 x2
mx n 3
y= 3+
for y to lie in [– 4, 3)
mx + n – 3 < 0
1 x2
x
R
this is possible only if m = 0
when, m = 0
then y = 3 +
note that n – 3 < 0 (think !)
n < 3 if
,
x
ymax
n 3
1 x2
3–
now
ymin occurs at x = 0 (as 1 + x2 is minimum)
ymin = 3 + n – 3 = n
n=–4
so m2 + n2 = 16
Ex.10 Find the domain and range of f(x) = sin
Sol.
4
2
x2 is positive and x < 4
1 – x should also be positive.
Thus the domain of n
n
4 x2
1 x
4 x2
1 x
n
4 x2
1 x
–2 < x < 2
x<1
is –2 < x < 1 sine being defined for all values, the domain of sin
is the same as the domain of n
To study the range. Consider the function
4 x2
1 x
4 x2
1 x
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FUNCTION
Page # 163
4 x2
4 x2
varies in the open interval (0, ) and hence n
varies from
1 x
1 x
As x varies from –2 to 1,
– to +
. Therefore the range of sin
Ex.11 Find the range of the function f(x) = sin–1
Sol.
Consider g(x) =
4 x2
1 x
n
is (–1, +1)
1 x4
.
1 5 x 10
1 x4
. Also g(x) is positive
1 5 x 10
x
R and g(x) is continuous
x
R and
g(0) = 1 and xlim g(x) = 0
Range of f(x) = sin–1 (g(x)) is 0,
g(x) can take all values from (0, 1]
Ex.12 f(x) = cos–1 {log [ [ x 3
2
.
1] ]}, find the domain and range of f(x (where [ * ] denotes the greatest
integer function).
Sol.
If cos–1 x = , then– 1
[ [x 3
0.37
1
1] ]
[x3] + 1 < 9
x
1
2.7
0
Domain of f(x) = Dr in x
Then 1
x3 + 1 < 9
Case I : 1
1
–1
log [ [x 3
1
[ x 3 1] < 3
1
1] ]
1
0 x<2
[0, 2)
Range of f(x) When 0
[x 3 1] < 2 then [ [ x3
8
1] ] = 1
[ x3
1
1] ]
2 2
[ x 3 1]
1
[x 3
1]
2.8 then [ [ x3
Rf is ( /2, cos–1 (log 2))
Ex.13 Find the range of the following functions
(i)
f(x) = loge (sinx sinx + 1) where 0 < x < /2.
(ii)
f(x) = loge (2 sin x + tan x – 3x + 1) where
6
x
e
x<2
1]
Case II : 2
Range in cos–1 {log 1} and cos–1 {log 2}
[ [x 3
[x3 + 1] < 9
[x3] < 8
[x3 + 1]
e–1
3
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2.8
1] ] = 2
FUNCTION
Page # 164
Sol.
(i)
0 < x < /2
0 < sin x < 1
Range of loge (sin x sin x + 1) for 0 < x < /2 = Range of loge (xx + 1) for 0 < x < 1
Let h(x) = xx + 1 = e
x log e x
+1
x log x
h (x) = e e (1 + loge x)
h (x) > 0 for x > 1/e and h (x) < 0 for x < 1/e
h(x) has a minima at x = 1/e
lim
x
Also xlim0 h(x) = 1 + e
0
ln x
1/ x
1
1+
e
0 < x< 1
loge
1
lim
=1+ e
1
e
x
0
1
e
1/ x
1 / x2
= 1 + e0 = 2 and xlim1 h(x) = 2
< (x x + 1) < 2
Y
1
e
< loge (x x +1) < loge 2
(1, 2)
(0, 2)
f(x)
1 + (1/e)1/e
Range of f(x) = loge 1 e
(ii)
1
e
O
2 cos3 x 3 cos2 x 1
cos2 x
(cos x – 1)2 cos x
h( /6)
h(x)
x
2 cos3 x – 3 cos2 x + 1 > 0
[ /6, /3]
h(x) is an increasing function of x
loge 2
h( /3)
Range of f(x) is loge 2
1
3
X
h (x) = (2 cos x + sec2 x – 3)
h (x) > 0
1
>0
2
1
, loge 2
Let h(x) = (2 sin x + tan x – 3x +1)
=
1/e
2
,log(1 2 3
1
2
3
loge h(x)
loge (1 + 2 3 – )
)
(11) Equal or Identical Functions : Two functions f & g are said to be equal if :
(i)
The domain of f = The domain of g
(ii)
The range of f = The range of g
(iii)
f(x) = g(x),
x
Df = Dg
Rf = R g
their common domain.
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FUNCTION
Page # 165
Ex.14 Let f(x) =
x
1
and g(x) = 2 then Df : R – {0} and Dg : R – {0}
x
x
Df = Dg
Hence both functions are identical
Ex.15 Let f(x) = sin x and g(x) =
Df
D.
Dg
1
then Df : x
cosec x
R and Dg : x
R – {n }
Hence both functions are non-identical
CLASSIFICATION OF FUNCTIONS
(1) One
One Function (Injective mapping) : A function f : A
B is said to be a one one
function or injective mapping if different elements of A have different f images in B. Thus for x1,
x2
A & f(x1) , f(x2)
B , f(x1) = f(x 2)
x1 = x2
or x1
x2
f(x1)
f(x 2).
Diagrammatically an injective mapping can be shown as
Remark :
(i) Any function which is entirely increasing or decreasing in its domain, is one one .
(ii) If any line parallel to x axis cuts the graph of the function atmost at one point,
then the function is one one.
(2) Many One function : A function f : A
B is said to be a many one function if two or more
elements of A have the same f image in B . Thus f : A
x1, x 2
A , f(x 1) = f(x2) but x1
B is many one if for
x2 .
Diagrammatically a many one mapping can be shown as
Remark :
(i)
A continuous function f(x) which has atleast one local maximum or local minimum, is
many one. In other words, if a line parallel to x axis cuts the graph of the function
atleast at two points, then f is many one.
(ii)
If a function is one one, it cannot be many one and vice versa.
(iii)
If f and g both are one-one, then fog and gof would also be one-one (if they exist).
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FUNCTION
Page # 166
x 2 8x 18
is not one-one.
x2 4 x 30
Ex.16 Show that the function f(x) =
Sol.
Test for one-one function
A function is one-one if f(x1) = f(x2)
Now f(x1) = f(x2)
x1 = x 2
x12
8 x1 18
x 22
8x 2
18
2
x1
4x1
x2
2
4x 2
30
12 x12 x 2 12x1x22
30
12x12 12 x22
312x1 312x 2
0
(x 1 – x2) {12x1x2 + 12(x1 + x2) – 312} = 0
x1 = x2 or x1 =
26 x2
1 x2
Since f(x1) = f(x2) does not imply x1 = x 2 alone, f(x) is not a one-one function.
Ex.17 Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy)
If f(4) = 65 and f(0)
Sol.
2, then show that f(x) – 1= x3
x
x, y
R.
R.
Given that f(x) f(y) + 2 = f(x) + f(y) + f(xy)
....(i)
Putting x = y = 0 in equation (i), we get f(0) f(0) + 2 = f(0) + f(0) + f(0)
or
(f(0))2 + 2 = 3f(0)
or
(f(0) – 2) (f(0) – 1) = 0 or f(0) = 1 (
f(0)
2) ....(ii)
Again putting x = y = 1 in equation (i) and repeating the above steps, we get
(f(1) – 2) (f(1) – 1) = 0
But f(1)
1 as f(x) is injective.
f(1) = 2
....(iii)
Now putting y = 1/x in equation (i), we get
f(x) f
1
1
x + 2 = f(x) + f x + f(1)
or
f(x) f
or
f(x)
Let
f(x) – 1 = g(x)
or
1
1
= f(x) + f
x
x
f
1
x
1
1. f
1
x
1 =1
or
f(x) f
1
1
x + 2 = f(x) + f x + 2
or
f(x) f
1
1
– f(x) – f
–1+ 1=0
x
x
or
{f(x) – 1}
f
g(x) g
g(x) = ± xn
f(x) = ± xn + 1
n= 3
1
x
1 =1
...(iv)
1
1
–1= g
x
x
1
= 1 which is only possible when
x
from equation (iv), we get
4n = 64 = (4)3
f
or f(x) = ± xn + 1 or65 = ± 4n + 1
f(x) = x3 + 1 or f(x) – 1 = x3 (neglecting negative sign)
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FUNCTION
Page # 167
(3) Onto-function (Surjective mapping) : If the function f : A
B is such that each element in
B (co domain) is the f image of atleast one element in A, then we say that f is a function of A
'onto' B . Thus f : A
B is surjective iff
b
B,
some a
A such that f (a) = b.
Diagramatically surjective mapping can be shown as
Note that :
if range
(4) Into function : If f : A
co domain, then f(x) is onto.
B is such that there exists atleast one element in co domain which is
not the image of any element in domain, then f(x) is into.
Diagramatically into function can be shown as
Remark :
(i)
If a function is onto, it cannot be into and vice versa .
(ii)
If f and g are both onto, then gof or fog may or may not be onto.
Thus a function can be one of these four types :
(a)
one one onto (injective & surjective)
(b)
one one into (injective but not surjective)
(c)
many one onto (surjective but not injective)
(d)
many one into (neither surjective nor injective)
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FUNCTION
Page # 168
Remark :
(i) If f is both injective & surjective, then it is called a Bijective function. Bijective functions are
also named as invertible, non singular or biuniform functions.
(ii) If a set A contains n distinct elements then the number of different functions defined from A
A
is nn & out of it n ! are one-one.
(iii) The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is
defined, then gof is also a bijection.
Ex.18 A function is defined as , f : D
R
f (x) = cot
1
(sgn x) + sin
1
(x
{x}) (where {x} denotes
the fractional part function) Find the largest domain and range of the function. State with reasons
whether the function is injective or not . Also draw the graph of the function.
Sol.
D [-1 , 2) ,
3
4 2 4
,
R =
,
f is many one
Ex.19 Find the linear function(s) which map the interval [ 0 , 2 ] onto [ 1 , 4 ].
Sol.
Let f (x) = a x + b
f (0) = 1 &
or
f (0) = 4 &
Ans. : f (x) =
Ex.20 (i)
f (2) = 4
b = 1
f (2) = 1
&
a =
b = 4
3x
+ 1 or f (x) = 4
2
&
3
2
a =
3
2
3x
2
Find whether f(x) = x + cos x is one–one.
(ii)
Identify whether the function f(x) = –x3 + 3x 2 – 2x + 4 ; R
(iii)
f(x) = x – 2x + 3; [0, 3]
2
R is ONTO or INTO
A. Find whether f(x) is injective or not. Also find the set A, if
f(x) is surjective.
Sol.
(i)
The domain of f(x) is R.
f (x)
0
x
f (x) = 1 – sin x.
complete domain
and equality holds at discrete points only
f(x) is strictly increasing on R. Hence f(x) is one-one.
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FUNCTION
Page # 169
(ii)
As codomain
range, therefore given function is ONTO
(iii)
f (x) = 2(x – 1); 0
x
ve ; 0
f (x) =
f(x)
6
3
x
1
ve ; 1 x
3
3
2
f(x) is a non monotonic continuous function.
Hence it is not injective.
1
0
3
x
For f(x) to be surjective, A should be equal to its range.
From graph,, range is [2, 6]
A
[2, 6]
Ex.21 If f and g be two linear functions from [–1, 1] onto [0, 2] and
(x) =
Sol.
f ( x)
g( x) , then show that
1
2
( ( x ))
: R+ – {–1, 1}
R be defined by
2.
Let h be a linear function from [–1, 1] onto [0, 2].
Let h(x) = ax + b, then h (x) = a
If a > 0, then h(x) is an increasing function & h(–1) = 0 and h(1) = 2
–a + b = 0 and a + b = 2
a = 1 & b = 1. Hence h(x) = x + 1.
If a < 0, then h(x) is a decreasing function & h(–1) = 2 and h(1) = 0
–a + b = 2 and a + b = 0
a = –1 & b = 1. Hence h(x) = 1 – x
Now according to the question f(x) = 1 + x & g(x) = 1 – x
or
f(x) = 1 – x
&
g(x) = 1 + x
Case-I :
When (x) =
1 x
1 x
Case-II :
When (x) =
1 x
,x
1 x
,x
–1
(x) =
f ( x)
g( x)
1 x
1 x
1 x or 1 x
1
x
;
1
x
1
In both cases, | (f(x)) + ( (1/x)) | = x
1
x
1
(where x > 0) =
x
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x
= – x.
1
x
2
2
2
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FUNCTION
Page # 170
E.
FUNCTIONAL EQUATION
Functional Equation is an equation where the unknown is a function. On solving such an equation we
obtain one or more functions as solutions. If x, y are independent variables, then :
(i)
f(xy) = f(x) + f(y)
f(x) = k ln x or f(x) = 0 .
(ii)
f(xy) = f(x) . f(y)
f(x) = xn ,
n
R
kx
(iii)
f(x + y) = f(x) . f(y)
f(x) = a , a > 0
(iv)
f(x + y) = f(x) + f(y)
f(x) = kx, where k is a constant.
Ex.22 (a) If f(x + y + 1) = ( f ( x)
f ( y) )2 and f(0) = 1
Sol.
f ( y) )2
Given f(x + y + 1) = ( f ( x)
Putting x = y = 0;
then f(1) = ( f (0)
Again putting x = 0, y = 1
(b)
Let f : R – {2}
2f(x) + 3f
Sol.
We have,
29
2
f
f (0) )2 = (1 + 2)2 = 32
f (1) )2 = (2 + 2)2 = 42
Similarly,
f(x) = (x + 1)2
R function satisfying the following functional equation,
2x 29
= 100x + 80,
x 2
f(x) = –
Replacing x by
2x
f
x
R. Determine f(x).
f (0) )2 = (1 + 1)2 = 22
Then f(2) = ( f (0)
and for x = 1, y = 1; f(3) = ( f (1)
x, y
x
R – {2}. Determine f(x).
2x 29
3
f
+ 50x + 40
x 2
2
...(i)
2x 29
in the given functional equation we get,
x 2
3
f
2
2x 29
x 2
2
2x
x
2x
x
29
2
29
2
29
50
2
3
2x 29
f ( x) 50
2
x 2
2x
x
29
+ 40
2
40
...(ii)
putting (ii) in (i), we get,
f(x) =
2x 29
9
f(x) – 75
– 60 + 50x + 40
x 2
4
2x 29
9
f(x) – f(x) = 20 – 50 x + 75
x 2
4
2x 29
5
f(x) = 20 – 50x + 75
x 2
4
f(x) = 16 – 40x + 60
( 2x
(x
29 )
2)
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FUNCTION
Page # 171
Ex.23 Let f be a function from the set of positive integers to the set of real numbers i.e., f : N R such that
(i) f(1) = 1; (ii) f(1) + 2f(2) + 3f(3) + ... + nf(n) = n (n + 1) f(n) for n 2 then find the value of f (1994).
Sol.
Given f(1) + 2f (2) + 3f(3) + ... + nf(n) = n(n + 1) f(n)
...(1)
Replacing n by (n + 1) then
f(1) + 2f(2) + 3f(3) + .... + nf(n) = n (n + 1) f(n + 1) = (n +1) (n + 2) f(n + 1) ...(2)
Subtracting (1) from (2) then we get
(n + 1) f(n + 1) = (n + 1) (n + 2) f (n + 1) – n (n + 1) f(n)
From which we conclude that
nf(n) = (n + 1) f(n + 1)
2f(2) = 3f (3) = 4f (4) = ... = nf(n)
Substituting the value of 2f(2), 3f(3), .... in terms of nf(n) in (1), we have
f(1) + (n – 1) nf (n) = n(n + 1) f(n)
f(1994) =
F.
f(1) = 2n f(n)
f(n) =
f (1)
2n
1
(
2n
f(1) = 1)
1
1
=
2.1994
3988
COMPOSITE FUNCTIONS
Let f: X
Y1 and g: Y2
Z be two functions and the set D = {x
X: f(x)
Y2}. If D
, then the
function h defined on D by h(x) = g{f(x)} is called composite function of g and f and is denoted by gof.
It is also called function of a function.
Remark : Domain of gof is D which is a subset of X (the domain of f). Range of gof is a subset of the
range of g. If D = X, then f(X) Y2.
Properties of composite functions :
(i) The composite of functions is not commutative i.e. gof
fog.
(ii) The composite of functions is associative i.e. if f, g, h are three functions such that fo (goh) &
(fog) oh are defined, then fo (goh) = (fog) oh.
Ex.24 Let f(x) = ex ; R+
Sol.
, . Find domain and range of fog (x)
2 2
R and g(x) = sin–1 x; [–1, 1]
Domain of f(x) : (0,
), Range of g(x) :
,
2 2
The values in range of g(x) which are accepted by f(x) are 0,
0 < g(x)
0 < sin–1 x
2
Hence domain of fog(x) is x
g
(0, 1]
0< x
2
1
(0, 1]
(0, /2]
f
0
/2
(e , e ]
Range
Domain
sin–1x
Therefore
2
Domain :
ex
(0, 1], Range :
(1/ e /2]
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FUNCTION
Page # 172
Ex.25 Let f(x) =
Sol.
x 1 2
, f (x) = f{f(x)}, f3 (x) = f{f2(x)},.....fk + 1 (x) = f{fk(x)}. for k = 1, 2, 3,...., Find f1998 (x).
x 1
f 1
x 1
f(x) =
, f2 (x) = f{f(x)} = f 1
x 1
f 3 ( x) 1
4
3
f = f{f (x)} = f 3 ( x) 1
x
x
x
x
x
x
x
x
1
1
1
1
1
1
1
x ,
1
1
1
= x,
1
1
1
1
1
x
1
1
x
f 2 ( x) 1
3
f (x) = f{f(x)} = f 2 ( x) 1
x 1
x 1,
x 1
= f(x)
x 1
f5(x) = f{f4 (x)} =
Thus, we can see that fk(x) repeats itself at intervals of k = 4.
Hence, we have f1998(x) = f2(x) =
Ex.26 Let g : R
1
x
[
1998 = 499 × 4 + 2]
R be given by g(x) = 3 + 4x. If gn(x) = gogo....og(x), show that fn(x) = (4n – 1) + 4nx if
g–n (x) denotes the inverse of gn (x).
Sol.
Since g(x) = 3 + 4x
g2(x) = (gog) (x) = g {g (x)} = g (3 + 4x) = 3 + 4 (3 + 4x) or g2(x) = 15 + 42x = (42 – 1) + 42x
Now g3(x) = (gogog) x = g {g2 (x) } = g (15 + 42 x) = 3 + 4 (15 + 42 x) = 63 + 43 x = (43 –1) + 43x
Similarly we get gn(x) = (4n – 1) + 4nx
Now leg gn (x) = y
n
x = g–n(y)
n
y = (4 – 1) + 4 x or
n
...(1)
–n
x = (y + 1 – 4 )4
...(2)
From (1) and (2) we get g–n (y) = (y + 1 – 4n) 4–n. Hence g–n (x) = (x + 1 – 4n) 4–n
Ex.27 If f(x) = | |x – 3| – 2 | ;
Sol.
2
g(x) =
6
x
x
0
1 x 2
2 x 3
x
4 and g(x) = 4 – |2 – x| ; –1
1 g( x ) 0 g( x ) 1
g( x ) 1 1 g( x ) 3
fog(x) =
5 g( x ) 3
g( x )
4
=
x
3 then find fog(x).
1 g( x ) for no value
g( x ) 1
1 x 1
5 g( x )
1 x
3
g(x)
4
2
=
x 1
1 x
1
5 (2
x)
1 x
2
5 (6
x)
2
3
x
=
x
1
1 x
3
x
1 x
2
x
1
2
3
x
2
1
1
–1
1
2
3
x
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FUNCTION
Page # 173
Ex.28 Prove that f(n) = 1 – n is the only integer valued function defined on integers such that
(i) f(f(n)) = n for all n
Sol.
Z and
(ii) f(f(n + 2) + 2) = n for all n
Z and
(iii) f(0) = 1.
The function f(n) = 1 – n clearly satisfies conditions (i), (ii) and (iii). Conversely, suppose a function
f:Z
Z satisfies (i), (ii) and (iii). Applying f to (ii) we get, f(f(f(n + 2) + 2) ) ) = f(n)
and this gives because of (i),
for all n
f(n + 2) + 2 = f(n),
........(1)
Z. Now using (1) it is easy to prove by induction on n that for all n
Z,
f (0 ) n if n is even
f(n) =
f (1) 1 n if n is odd
Also by (iii), f(0) = 1. Hence by (i), f(1) = 0. Hence f(n) = 1 – n for all n
G.
Z.
GENERAL DEFINITION
(1) Identity function : A function f : A
denoted by IA. Ex : f : R
+
A defined by f(x) = x
R ; f(x) = e
+
nx
and f : R
x
A is called the identity of A &
R ; f(x) = n ex
Every Identity function is a bijection.
(2) Constant function : A function f : A
B is said to be constant function. If every element of set
A has the same functional image in set B i.e. f : A
B ; f(x) = c
x
A&c
B is called constant
function.
(3) Homogeneous function : A function is said to be homogeneous w.r.t. any set of variables when
each of its term is of the same degree w.r.t. those variables.
(4) Bounded Function : A function y = f(x) is said to be bounded if it can be express is the form of
a
f(x)
b where a and b are finite quantities.
Ex : –1
sin x
1 ; 0
{x} < 1 ; –1
sgn (x)
1 but ex is not bounded.
Ex : Any function having singleton range like constant function.
(5) Implicit function & Explicit function : If y has been expressed entirely in terms of ‘x’ then it is
called an explicit function.
If x & y are written together in the form of an equation then it is known as implicit equation
corresponding to each implicit equation there can be one, two or more explicit function satisfying it
Ex : y = x3 + 4x 2 + 5x
H.
Explicit function
Ex : x + y = 1
Implicit equation
Ex : y = 1 – x
Explicit function
EVEN & ODD FUNCTIONS
Function must be defined in symmetric interval [–x, x]
If f ( x) = f (x) for all x in the domain of ‘f’ then f is said to be an even function.
e.g. f (x) = cos x ; g (x) = x² + 3.
If f ( x) = f (x) for all x in the domain of ‘f’ then f is said to be an odd function.
e.g. f (x) = sin x ; g (x) = x3 + x.
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FUNCTION
Page # 174
Remark :
(a) f (x)
f ( x) = 0
f (x) is even & f (x) + f ( x) = 0
f (x) is odd .
(b) A function may be neither even nor odd.
(c) Inverse of an even function is not defined.
(d) Every even function is symmetric about the y axis & every odd function is symmetric about the
origin .
(e) A function (whose domain is symmetric about origin) can be expressed as a sum of an even & an
odd function. e.g. f ( x )
f ( x ) f ( x)
2
f ( x) f ( x)
2
EVEN
ODD
(f) The only function which is defined on the entire number line & is even and odd at the same time is
f(x) = 0 .
(g) If f and g both are even or both are odd then the function f.g will be even but if any one of them
is odd and other even then f.g will be odd.
Ex.29 Which of the following functions is odd ?
(A) sgn x + x2000
Sol.
(B) | x | – tan x
(C) x3 cot x
(D) cosec x 55
Let’s name the function of the parts (A), (B), (C) and (D) as f(x), g(x), h(x) & (x) respectively. Now
(A) f(–x) = sgn (–x) + (–x)2000 = –sgn x + x2000
f(x) &
–f(x)
f is neither even nor odd.
(B) g(–x) = |–x| – tan (–x) = |x| + tan x
g is neither even nor odd.
(C) h(–x) = (–x)3 cot (–x)
h is an even function
= –x 3 (–cot x) = x3 cot x = h(x)
(D) (–x) = cosec (–x)55 = cosec (–x55 )
= –cosec x 55 = – (x)
is an odd function.
Alternatively
(A)
f(x) = sgn (x) + x2000 = O + E = neither E nor O
(B)
g(x) = E – O = Neither E nor O
(C)
h(x) = O × O = E
(D)
f(–x) = O o O = O
Ex.30 f(x) = (tan x5) e x
Sol.
3
sgn x 7
(D) is the correct option
is
(A) an even function
(B) an odd function
(C) neither even nor odd function
(D) none of these
f(x) = (tan (x5)) e x
O
(O)
3
eO
sgn ( x 7 )
×
O
(O)
= O × e O × O = O × eE
=O×E=O
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FUNCTION
Page # 175
x tan x, 0 x
Ex.31 Let f: [–2, 2]
2
(i) f is an odd function
Since f(x) =
2
[x],
2
[x],
2
x
x
2
( x ) tan( x),
0
2
2
[ x ],
2 or f (–x) =
x 2
2
If f is an odd function then f(x) = –f (–x) =
2
If f is an even function
[ x],
f(x) = f(–x) =
2
x
2
2
x tan x,
(ii)
x tan x,
x
f(–x) =
x tan x,
(i)
[–2, 0] so that
(ii) f is an even function (where [*] denotes the greatest integer function)
x tan x, 0 x
Sol.
2 Define f for x
2
R be a function if f(x) =
[ x],
x
2
[ x],
x
2
2
0
x
2
0
2
x
2
2
x
0
2
Ex.32 Let f(x) = ex + sin x be defined on the interval [–4, 0]. Find the odd and even extension of f(x) in the
interval [–4, 4].
Sol.
Odd Extension : Let g 0 be the odd extension of f(x), then
g0 (x) =
f ( x)
; x [ 4,0]
=
f ( x ) ; x [0,4]
e x sin x ; x [ 4,0]
e x sin x ; x [0,4]
Even Extension : Let ge be the even extension of f(x), then
ge (x) =
I.
f ( x ) ; x [ 4,0]
e x sin x ; x [ 4,0]
=
f ( x ) ; x [0,4 ]
e x sin x ; x [0,4]
PERIODIC FUNCTION
A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the
function such that f (x + T) = f(x), for all values of x and x + T within the domain of f(x). The least
positive period is called the principal or fundamental period of f.
e.g. The function sin x & cos x both are periodic over 2 & tan x is periodic over .
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FUNCTION
Page # 176
Remark :
(a)
A constant function is always periodic, with no fundamental period.
(b)
If f(x) has a period p, then
(c)
if f(x) has a period T then f(ax + b) has a period T/a (a > 0).
(d)
If f(x) has a period T1 & g(x) also has a period T2 then period of f(x) ± g(x) or
1
and
f( x )
f (x) also has a period p.
f ( x)
g( x) is L.C.M of
T1 & T2 provided their L.C.M. exists. However that L.C.M. (if exists) need not to be fundamental
period. If L.C.M. does not exists then f(x) ± g(x) or f(x) . g(x) or
f ( x)
is nonperiodic e.g. |sin x|
g( x)
has the period , |cos x| also has the period
|sin x| + |cos x| also has a period . But the fundamental period of |sin x| + |cos x| is /2.
(e)
If g is a function such that gof is defined on the domain of f and f is periodic with T, then gof is
also periodic with T as one of its periods. Further if
# g is one-one, then T is the period of gof
# g is also periodic with T’ as the period and the range of f is a subset of [0, T’], then T is the
period of gof
(f)
Inverse of a periodic function does not exist.
Ex.33 Find period of the following functions
Sol.
x
x
+ cos
2
3
(i)
f(x) = sin
(ii)
f(x) = {x} + sin x
(iii)
f(x) = cos x . cos 3x
(iv)
f(x) = sin
(i)
Period of sin x/2 is 4 while period of cos x/3 is 6 . Hence period of sin x/2 + cos x/3 is 12
3x
x
2x
– cos
– tan
.
2
3
3
{L.C.M. or 4 & 6 is 12}
(ii)
Period of sin x = 2 ; Period of {x} = 1;
but L.C.M. of 2 & 1 is not possible
(iii)
f(x) = cos x . cos 3x ; Period of f(x) is L.C.M. of 2 ,
2
3
it is aperiodic
=2
but 2 may or may not be the fundamental period. The fundamental period can be
n
N. Hence cross-checking for n = 1, 2, 3..... we find
2
, where
n
to be fundamental period
f( + x) = (–cos x) (– cos 3x) = f(x)
(iv)
Period of f(x) is L.C.M. of
2
2
4
2
,
,
= L.C.M. of
,6 ,
= 12
3 / 2 1/ 3 3 / 2
3
3
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FUNCTION
Page # 177
Ex.34 If f(x) = sin x + cos ax is a periodic function, show that a is a rational number.
Sol.
Given f(x) = sin x + cos ax
2
2
and period of cos ax =
1
a
L.C.M. of { 2 , 2 }
2 2
,
Hence period of f(x) = L.C.M. or
=
1 a
H.C.F. of {1, a}
Period of sin x =
2
k
where k = H.C.F. of 1 and a
1
a
= integer = q (say), ( 0) and
= integer = p (say)
k
k
a/k
1/ k
p
q
a=
p
q
a = rational number
Ex.35 Given below is a partial graph of an even periodic function
f whose period is 8. If [*] denotes greatest integer function
then find the value of the expression.
f (–3) + 2 | f (–1) | +
Sol.
f (–3) = f (3) = 2
f
7
8
+ f (0) + arc cos (f(–2)) + f (–7) + f (20)
[ f (x) is an even function,
f (– x) = f (x) ]
again f (–1) = f (1) = – 3
2 | f (–1) | = 2 | f (1) | = 2 | – 3 | = 6
from the graph,
f (0) = 0
cos–1
–3< f
7
8
f
<–2
7
8
=– 3
(obviously from the graph)
f ( 2) = cos–1 f (2) = cos –1 (1) = 0
f (–7) = f (– 7 + 8) = f (1) = – 3
f (20) = f (4 + 16) = f (4) = 3
[f (x) has period 8]
[ f (nT + x) = f (x) ]
sum = 2 + 6 – 3 + 0 + 0 – 3 + 3 = 5
Ex.36 Check whether the function defined by f(x + ) = 1 +
2f ( x ) f 2 ( x )
x
R, is periodic or not, if
periodic, then find its period.
Sol.
The given function is true if 2f(x) – f2(x)
0
f(x)[f(x) – 2]
Also from the given function, it is clear that f(x + )
From (i) and (ii), we conclude that 1
f(x)
Replacing x by x +
2
0
0
f(x)
f(x)
2 ....(i)
1
{f(x + ) – 1}2 = – {(f(x) – 1)2} + 1
in above equation, we get {f(x + 2 ) – 1}2 = – {f(x + ) – 1}2 + 1
From (iv) – (iii), we get
{f(x + 2 ) – 1}2 = {f(x) – 1)}2
....(ii)
2
Again, we have {f(x + ) – 1} = 2f(x) – f (x)
2
1
f(x + 2 ) = f(x)
f is periodic function with period 2 .
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...(iii)
...(iv)
FUNCTION
Page # 178
Ex.37 If the periodic function f(x) satisfies the equation f(x + 1) + f(x –1) =
3 f(x)
x
R then find the
period of f(x)
Sol.
We have f(x + 1) + f(x – 1) =
3 f(x)
x
R
...(1)
Replacing x by x – 1 and x + 1 in (1) then f(x) + f(x – 2) =
and
f(x + 2) + f(x) =
3 f(x – 1)
3 f(x + 1)
...(3)
Adding (2) and (3), we get 2f(x) + f(x – 2) + f(x + 2) =
2f (x) + f(x – 2) + f(x + 2) = 3 .
...(2)
3 (f (x – 1) + f(x + 1))
3 f(x)
[From (1)]
f(x + 2) + f(x – 2) = f(x)
...(4)
Replacing x by x + 2 in equation (4) then f (x + 4) + f (x) = f (x + 2)
...(5)
Adding equations (4) and (5), we get f(x + 4) + f (x – 2) = 0
...(6)
Again replacing x by x + 6 in (6) then f (x + 10) + f (x + 4) = 0
...(7)
Subtracting (6) from (7), we get f (x + 10) – f (x – 2) = 0
...(8)
Replacing x by x + 2 in (8) then f (x + 12) – f(x) = 0 or f (x + 12) = f(x)
Hence f(x) is periodic function with period 12.
J.
INVERSE OF A FUNCTION
Let f : A
g: B
g=f
B be a one one & onto function, then their exists a unique function
A such that f(x) = y
1
: B
A = {(f(x), x)
g(y) = x,
(x, f(x))
x
A & y
B . Then g is said to be inverse of f. Thus
f}.
Properties of inverse function :
(i)
The inverse of a bijection is unique, and it is also a bijection.
(ii)
If f : A
B is a bijection & g : B
A is the inverse of f, then fog = IB and
gof = IA , where IA & IB are identity functions on the sets A & B respectively.
(iii)
The graphs of f & g are the mirror images of each other in the line y = x.
(iv)
Normally points of intersection of f and f–1 lie on the straight line y =x. However it must be noted
that f(x) and f–1(x) may intersect otherwise also.
(v)
In general fog(x) and gof(x) are not equal. But if either f and g are inverse of each other or
atleast one of f, g is an identity function, then gof = fog.
(vi)
If f & g are two bijections f : A
and (gof)
1
=f
1
B, g : B
C then the inverse of gof exists
o g 1.
Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)
FUNCTION
Page # 179
Ex.38 Find the inverse of the function f(x) = ln(x2 + 3x +1); x
[1, 3] and assuming it to be an onto function.
2x
Sol.
Given f(x) = ln (x2 + 3x + 1)
f (x) =
(x
2
3
>0
3 x 1)
x
[1, 3]
which is a strictly increasing function. Thus f(x) is injective, given that f(x) is onto. Hence the given
function f(x) is invertible. Now let y = f(x) = ln (x2 + 3x + 1) then x = f–1 (y)
and y = ln (x + 3x + 1)
e = x + 3x + 1
2
3
x=
9
4.1 .(1 e y )
2
From (1) and (2), we get
y
2
3
(5
2
3
f–1 (y) =
Given f(x) = f(x) =
x,
x
x2 ,
1 x
8 x,
x
(5
x + 3x + 1 – ey = 0
4e y )
x,
x
2
x ,
1 x
8 x,
x
(5
2
4e y )
(
x
[1, 3]) ...(2)
3
Hence f–1 (x) =
4e x )
(5
2
1
4
4
1
4
4
y,
Let f(x) = y
3
=
2
Ex.39 Find the inverse of the function f(x) =
Sol.
4ey )
...(1)
2
x = f–1(y) ....(1)
y
y,
x=
1
2
y
2
y,
1
y,
4 =
y / 64, y / 64
y
1 y
2
y / 64,
4
1
y
16
16
y,
y,
y 1
x,
x 1
–1
1 y 16 [From (1)]. Hence f (x) =
x,
1 x 16 .
f (y) =
y 2 / 64,
y 16
y 2 / 64,
x 16
–1
Ex.40 A function f : 3 ,
7,
4
2
defined as, f (x) = x2
solution of the equation, f (x) = f
Sol.
f (x) = y = x
2
x =
3
graphs of f
9
1
3x + 4
4 (4
2
y)
1
x
3
1
(x) and find the
(x) .
2
=
3 x + 4 . Then compute f
4y
2
3 x + (4
7
y) = 0
f
1
(y) = 3 +
(x) & f (x) intersect each other at y = x
4x 7
2
f (x) = x
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3x+y=x
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x=2
FUNCTION
Page # 180
EXERCISE – I
JEE MAIN
log0.3 ( x 1)
1. The domain of the function f(x)=
(A) (1, 4) (B) (–2, 4)
Sol.
x
(C) (2, 4)
2
2x
(D) [2,
8
is
Sol.
)
2. The domain of the function
f(x) = log1/2
log2 1
(A) 0 < x < 1 (B) 0 < x
Sol.
1
4
x
4. Find domain of the function
1 is
1 (C) x
1
(D) null set
f(x) =
log x
4
log2
2
2x 1
3 x
(A) (–4, –3) (4, )
(C) (– , – 4) (3, )
Sol.
(B) (– , –3) (4,
(D) None of these
2
3. If q – 4 p r = 0, p > 0, then the domain of the
3
2
function, f(x) = log (px + (p + q) x + (q + r) x + r) is
(A) R –
q
2p
(C) R– (
, 1]
(B) R – (
q
2p
, 1]
q
2p
(D) none of these
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)
FUNCTION
Page # 181
5. The domain of the function
log1 / 3 log4 ([x]2
5) is
f(x) = log
(where [x] denotes greatest integer function)
(A) [–3, –2) [3, 4)
(B) [–3, –2) (2, 3]
(C) R – [–2, 3)
(D) R – [–3, 3]
Sol.
6. Range of f(x) = 4 x + 2x + 1 is
(A) (0, ) (B) (1, )
(C) (2,
Sol.
)
(D) (3,
8. The range of the functin
2
2
( 2– log2 (16 sin x + 1)) is
(A) (– , 1) (B) (– , 2)
Sol.
(C) (– , 1]
)
cos
9. Range of the function f(x)=
x
2
1
cos
(A) [0, 2] (B) [0, 4]
Sol.
7. Range of f(x) = log
5
(D) (– , 2]
1
cos
x
2
(C) [2, 4]
1
x
2
1
3
2
(D) [1, 3]
(D) None of these
Sol.
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x
is
2
1
{ 2 (sin x –cos x) + 3} is
(A) [0, 1] (B) [0, 2] (C) 0,
cos
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FUNCTION
Page # 182
sin2 x 4 sin x 5
10. If f(x) =
, then range of f(x) is
2 sin2 x 8 sin x 8
(A)
1
,
2
(B)
5
,1
9
(C)
5
,1
9
(D)
5
,
9
12. Which of the following represents the graph of
f(x) = sgn ([x + 1])
(A)
–1
1
–1
(B)
1
–1
–1
Sol.
(C)
–1
1
–1
(D)
1
–1
1
1
–1
Sol.
11.
1
2
The sum
1
2
1
2000
1
2
2
2000
1
2
3
2000
......
1
2
1999
2000
is equal to
(where [ * ] denotes the greatest integer function)
(A) 1000 (B) 999
(C) 1001 (D) None of these
Sol.
13. If f(x)=2 sin2 +4 cos (x+ ) sin x. sin +cos (2x+2 )
then value of f2(x) + f2
(A) 0
Sol.
(B) 1
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4
x is
(C) –1
(D) x 2
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FUNCTION
Page # 183
14. Let f(x) = ax 2 + bx + c, where a, b, c are rational
and f : Z
Z, where Z is the set of integers. Then
a + b is
(A) a negative integer
(B) an integer
(C) non-integral rational number (D) None of these
Sol.
Sol.
17. Let f : R
f(x) =
15. Which one of the following pair of functions are
identical ?
(A) e
(ln x)/2
–1
and
R be a function defined by
2x 2 x 5
then f is
7x 2 2x 10
(A) one – one but not onto
(B) onto but not one – one
(C) onto as well as one – one
(D) neither onto nor one – one
Sol.
x
–1
(B) tan (tan x) & cot (cot x)
2
4
2
4
(C) cos x + sin x and sin x + cos x
(D)
|x|
and sgn (x) where sgn(x) stands for signum
x
function.
Sol.
18. Let f : R
R be a function defined by
3
2
f(x) = x + x + 3x + sin x. Then f is
(A) one – one & onto
(B) one – one & into
(C) many one & onto
(D) many one & into
Sol.
16. The function f : [2, )
Y defined by
2
f(x) = x – 4x + 5 is both one–one & onto if
(A) Y = R (B) Y = [1, ) (C) Y = [4, ) (D) Y = [5, )
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FUNCTION
19. If f(x) =
Page # 184
4a 7 3
2
x + (a – 3) x + x + 5 is a one–
3
one function, then
(A) 2 a 8
(C) 0 a 1
Sol.
Sol.
(B) 1 a 2
(D) None of these
22. Let ‘f’ be a function from R to R given by
f(x) =
x2 4
. Then f(x) is
x2 1
(A) one-one and into
(C) many-one and into
Sol.
20. Let f: (e, ) R be defined by f(x) = ln (ln(ln x)),
then
(A) f is one one but not onto
(B) f is onto but not one – one
(C) f is one–one and onto
(D) f is neither one–one nor onto
Sol.
–1
23. If f(x) = cot
(B) one-one and onto
(D) many-one and onto
0,
+
x:R
2
2
and g(x) = 2x – x : R
R. Then the range of the
function f(g(x)) wherever define is
(A) 0,
2
(B) 0,
4
(C)
,
4 2
(D)
Sol.
21. The function f : R
(A) one-one and onto
(C) one-one and into
R defined by f(x) = 6x + 6|x| is
(B) many-one and onto
(D) many-one and into
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4
FUNCTION
Page # 185
1 if
24. Let g(x) = 1 + x – [x] and f(x) =
0
1
if
if
x
0
x
x
0
,
0
Sol.
then
x, fog(x) equals
(where [ * ] represents greatest integer function).
(A) x
(B) 1
(C) f(x)
(D) g(x)
Sol.
27.
If y = f (x) satisfies the condition
f x
1
1
2
=x + 2 (x
x
x
2
25. Let f: [0, 1]
[1, 2] defined as f(x) = 1 + x and
g : [1, 2]
[0, 1] defined as g(x) = 2 – x then the
composite function gof is
(A) injective as well as surjective
(B) Surjective but not injective
(C) Injective but non surjective
(D) Neither injective nor surjective
Sol.
(A) – x + 2
Sol.
0) then f(x) equals
2
(B) – x – 2
28. The function f(x) = log
(A) even
(C) neither even nor odd
Sol.
2
(C) x + 2
1 sin x
is
1 sin x
(B) odd
(D) both even & odd
26. Let f & g be two functions both being defined
from R
R as follows f(x) =
x
g(x) =
x
2
for
x
0
for
x
0
x |x|
and
2
. Then
(A) fog is defined but gof is not
(B) gof is defined but fog is not
(C) both fog & gof are defined but they are unequal
(D) both gof & fog are defined and they are equal
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2
(D) x – 2
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FUNCTION
Page # 186
29. It is given that f(x) is an even function and satisfy the
2
relation f(x) =
(A) 10
Sol.
xf ( x )
then the value of f(10) is
2 tan 2 x.f ( x 2 )
(B) 100
(C) 50
(D) None of these
32. Let f(x) = x (2 – x), 0 x 2. If the definition of ‘f’
is extended over the set, R – [0, 2] by f(x + 2) = f(x),
then ‘f’ is a
(A) periodic function of period 1
(B) non-periodic function
(C) periodic function of period 2
(D) periodic function of period 1/2
Sol.
30. Fundamental period of f(x) = sec (sin x) is
(A) /2
(B) 2
(C)
(D) a periodic
Sol.
33. Let f(2, 4)
f(x) = x –
(1, 3) be a function defined by
x
–1
, then f (x) is equal to
2
(where [ * ] denotes the greatest integer function)
x
31. The period of sin
[x] + cos
+ cos
[x],
4
2
3
(A) 2x
(B) x +
x
2
(C) x + 1
(D) x – 1
Sol.
where [x] denotes the integral part of x is
(A) 8
(B) 12
(C) 24
(D) Non–periodic
Sol.
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FUNCTION
Page # 187
EXERCISE – II
JEE ADVANCED (OBJECTIVE )
LEVEL – I
SINGLE CORRECT
1. If domain of f(x) is (– , 0] then domain of
2
f(6{x} – 5{x} + 1) is
(where {*} represetns fractional part function)
n
(A)
n
n
(C)
n
1
,n
3
1
2
1
n 1
6
(B) (– , 0)
3. If [2 cos x] + [sin x] = –3, then the range of the
function, f(x) = sin x +
3 cos x in [0, 2 ] is
(where [ * ] dentoes greatest integer function)
(A) [–2, –1)
(B) (–2, –1]
(C) (–2, –1)
(D) [–2, –
3)
Sol.
(D) None of these
Sol.
2.
f : R – {3}
R – {1} defined by f(x)=
3x 2
x 1
(B)
x 1
3x 2
(D) Does not exist
(a)
x 2 –1
. f is equal to
x 3
(C)
x 2
x 3
Sol.
4. In the square ABCD with side AB = 2, two points
M & N are on the adjacent sides of the square such
that MN is parallel to the diagonal BD. If x is the
d i s t a n c e of M N f ro m t h e ve rt e x A a n d
f(x) = Area ( AMN), then range of f(x) is
(A) (0, 2 ] (B) (0, 2]
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(C) (0,2 2 ]
(D) (0, 2 3 ]
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FUNCTION
Page # 188
Sol.
Sol.
7. If f(x) = 2log10 x + 8, then solution of equation f(x)
= f–1 (x)
(A) 1 (B) 10
5. Let f be a real valued function defined by
f(x) =
(A) R
(C) 100 (D)
1
10
e x e |x|
then the range of f(x) is
e x e|x|
(B) [0, 1]
(C) [0, 1)
(D) 0,
1
2
Sol.
8. If A, B, C are three decimal numbers and
p = [A + B + C] and q = [A] + [B] + [C] then maximum
value of p – q is (where [ * ] represents greatest
integer function).
(A) 0
(B) 1
(C) 2
(D) 3
Sol.
6. The number of solution(s) of the equation
[x] + 2{–x} = 3x, is/are
(where [ * ] represents the greatest integer function
and { * } denotes the fractional part of x)
(A) 1
(B) 2
(C) 3
(D) 0
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FUNCTION
Page # 189
9. If f(x) = 2[x] + cos x, then f: R
R is
(where [ * ] denotes greatest integer function)
(A) one–one and onto
(B) one–one and into
(C) many–one and into (D) many–one and onto
Sol.
Sol.
12. Function f : (– , 1)
(0, e5] defined by
( x2 3 x 2 )
f(x) = e
is
(A) many one and onto
(C) one one and onto
Sol.
(B) many one and into
(D) one one and into
10. If the real-valued function f(x) = px + sinx is a
bijective function, then the set of all possible values
of p R is
(A) R – {0} (B) R
(C) (0, ) (D) None of these
Sol.
+
x
13. f(x) = |x – 1|, f : R
R ; g(x) = e , g : [–1, ) R
If the function fog(x) is defined, then its domain and
range respectively are
(A) (0, ) & [0, )
(B) [–1, ) & [0, )
(C) [–1,
)& 1
1
,
e
(D) [–1,
)&
1
1,
e
Sol.
11. Let S be the set of all triangles and R+ be the set
of positive real numbers. Then the function, f : S R+,
f( ) = area of the , where
S is
(A) injective but not surjective
(B) surjective but not injective
(C) injective as well as surjective
(D) neither injective nor surjective
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FUNCTION
Page # 190
14. If f(1) = 1 and f(n + 1) = 2f(n) + 1 if n 1, then
f(n) is equal to
n
n
n
n–1
(A) 2 + 1
(B) 2
(C) 2 – 1
(D) 2
–1
Sol.
15. A function f : R
R satisfies the condition,
2
4
x f(x) + f(1 – x) = 2x – x . Then f(x) is
2
2
2
4
(A) – x – 1 (B) –x + 1 (C) x – 1 (D) – x + 1
Sol.
16. If the graph of the function f(x) =
Sol.
18. If f(x) = sin
[a] x has as its fundamental period
then (where [ * ] denotes the gratest integer function)
(A) a = 1 (B) a = 9 (C) a [1, 2) (D) a [4, 5)
Sol.
ax 1
is
x (a x 1)
n
symmetric about y–axis, then n is equal to
(A) 2
(B) 2/3
(C) 1/4
(D) –1/3
Sol.
19. The fundamental period of function
f(x) = [x] + x
(A) 1/3
Sol.
17.
If g : [–2, 2]
R where g(x)=x3 +tan x +
1
3
(B) 2/3
x
2
– 3x + 15
3
(C) 1
(D) Non–periodic
x2 1
p
be an odd function , then the value of the parameter
P is
(A) –5 < P < 5 (B) P < 5 (C) P>5 (D) None of these
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FUNCTION
Page # 191
LEVEL – II
MULTIPLE CORRECT
1. Let f : [–1, 1] [0, 2] be a linear function which is
onto then f(x) is/are
(A) 1 – x (B) 1 + x
(C) x – 1
(D) x + 2
Sol.
3. A function ‘f’ from the set of natural numbers to
n 1
, when n is odd
2
integers defined by f(n)=
is.
n
, when n is even
2
(A) one–one
Sol.
(B) many–one
(C) onto
(D) into
2. In the following functions defined from [–1, 1] to
[–1, 1] the functions which are not bijective are
(A) sin (sin
–1
x)
(C) (sgn x ) ln e
Sol.
(B)
x
2
–1
sin
(sin x)
3
(D) x sgn x
4. Let f(x) =
0
x
1 x
,0 x
1 x
1 and g(x) = 4x (1 – x),
1. then
1 4x 4 x 2
,0 x 1
1 4x 4 x 2
1 4x 4 x 2 1
,
(B) fog =
x 1
1 4x 4x 2 2
8 x(1 x )
(C) gof =
,0 x 1
(1 x)2
8 x(1 x)
(D) gof =
,0 x 1
(1 x) 2
(A) fog =
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FUNCTION
Page # 192
Sol.
7. If f : R
[–1, 1], where f(x) = sin /2 [x],
(where [*] dentoes the greatest integer function) then
(A) f(x) is onto
(B) f(x) is into
(C) f(x) is periodic
(D) f(x) is many one
Sol.
5. If ‘f’ and ‘g’ are bijective functions and gof is
defined then gof must be
(A) injective
(B) surjective
(C) bijective
(D) into only
Sol.
4
4
6. The period of the function f(x) = sin 3x + cos 3x is
(A) /6
(B) /3
(C) /2
(D) /12
Sol.
8. If F(x) =
sin [ x]
{x } , then F(x) is
(where {*} denotes fractional part of function and
[*] denotes greatest integer function)
(A) periodic with fundamental period 1
(B) even
(C) range is singleton
(D) identical to sgn
sgn
{ x}
{ x}
– 1,
(where { * } denotes fractional part of function
and [ * ] denotes greatest integer function and sgn
(x) is a signum function)
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FUNCTION
Page # 193
Sol.
10. Which of the following functions are periodic ?
(A) f(x) = sgn (e–x )
(B) f(x) =
1 if x is a rational number
0 if x is an irrational number
8
8
1 cos x 1 cos x
1
1
x
(D) f(x) = x
+ 2 [–x]
2
2
(C) f(x) =
(where [ * ] denotes greatest integer function)
Sol.
2
9. Function f(x) = sin x + tan x + sgn (x – 6x + 10) is
(A) periodic with period 2
(B) periodic with period
(C) Non–periodic
(D) periodic with period 4
Sol.
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FUNCTION
Page # 194
EXERCISE – III
JEE ADVANCED
Matrix Match Type
Comprehension
Consider the function f(x) =
x2 1,
nx,
1. The graph of the function y = f(x) is as follows.
1 x 1
1 x e
Let f1 (x) = f(|x|)
f2 (x) = |f(|x|)|
f3 (x) = f(–x)
Now answer the following questions.
1. Number of positive solution of the equation
2f2 (x) – 1 = 0 is (are)
(A) 4
(B) 3
(C) 2
(D) 1
Sol.
y
1
–2
–1
O
1
–1
2
x
Match the function mentioned in Column-I with the
respective graph given in Column-II.
1. Column – I
Column – II
y
1
(A) y = |f (x) |
(P)
–2
–1
1
2
O
x
–1
y
1
(B) y = f(|x|)
(Q) –2
–1 O
x
2
1
–1
y
(C) y = f(–|x|)
(R)
1
–2 –1
O
1
2
1
2
x
–1
2. Number of integral solution of the equation
f1 (x) = f2 (x) is (are)
(A) 1
(B) 2
(C) 3
(D) 4
Sol.
y
(D) y =
1
(|f(x)| – f(x))
2
1
(S)
–2 –1
O
–1
Sol.
3. If f4 (x) = log27 (f3(x) + 2), then range of f4(x) is
(A) [1, 9] (B)
1
,
3
(C) 0,
1
3
(D) [1, 27]
Sol.
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x
FUNCTION
Page # 195
Sol.
2. Column - I contains functions and column II
contains their natural domains. Exactly one entry of
column II matches with exactly one entry of column I.
Column – I
Column – II
–1
(A) f(x) = sin
(B) g(x) =
n
x 1
x
x2
(P) (1, 3)
3x 2
x 1
(x) = n
x2
(R)
12
)
(Q) (– , 2)
1
x
1
(C) h(x) = n
2
(D)
(3,
2x
,
1
2
(S) [–3, –1) [1, ¥)
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FUNCTION
Page # 196
Subjective Type
1. Find the domain of each of the following functions
(i) f(x) =
x3
5x 3
x2 1
1
x 1/ 2
(v) logx log2
Sol.
Sol.
1
(ii) f(x) =
x |x |
Sol.
(vi) f(x) =
3 2x
21
x
Sol.
x + sin x
(iii) f(x) = e
Sol.
(iv) f(x) =
Sol.
1
log10 (1 x )
x 2
(vii) f(x) =
1
1 x2
Sol.
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FUNCTION
Page # 197
2
–3/2
(viii) f(x) = (x + x + 1)
Sol.
(xii) f(x) =
log1/ 4
5x x 2
4
Sol.
(ix) f(x) =
x 2
x 2
1 x
1 x
Sol.
2
(xiii) f(x) = log10 (1 – log10 (x – 5x + 16))
Sol.
(x) f(x) =
tan x
tan 2 x
Sol.
2. Find the domain of definitions of the functions
(Read the symbols [*] and {*} as greatest integers
and fractional part functions respectively.)
(i) f(x) =
cos 2 x
16
x2
Sol.
(xi) f(x) =
1
1 cos x
Sol.
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FUNCTION
Page # 198
3
2
(ii) f(x) = log7 log5 log3 log2 (2x + 5x – 14x)
Sol.
(vi) f(x) = log100x
2 log10 x 1
x
Sol.
(iii) f(x) = ln ( x 2
5x
24 – x – 2)
Sol.
(iv) f(x) =
1
(vii) f(x) =
2
+ ln x(x – 1)
4x 2 1
Sol.
1 5x
7 x 7
Sol.
(viii) f(x) =
log 1
2
x
x
2
1
Sol.
(v) y = log10 sin (x – 3) +
16
x2
Sol.
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FUNCTION
(ix) f(x) =
Page # 199
x2 | x |
1
9 x2
Sol.
cos x
(xii) f(x) =
6
35 x
1
2
6x
2
Sol.
(x) f(x) =
(x 2
3x 10 ). n2 ( x 3)
Sol.
(xiii) f(x) =
(xi) f(x) =
log x (cos 2 x)
1
2
+ log(2{x}–5) (x – 3x + 10) +
[ x]
Sol.
Sol.
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1
1 |x |
FUNCTION
Page # 200
(xiv) f(x) = logx sin x
Sol.
(xvi) f(x) =
1
1
1
2
+log1 – {x}(x –3x+10)+
+
[ x]
sec(sin x )
2 |x|
Sol.
1
+
x
sin
100
log10 (log10 x ) log10 ( 4 log10 x) log10 3
(xv) f(x) = log2
log1 / 2 1
Sol.
(xvii) f(x) =
(5x 6 x2 )[{ n{ x}}] +
(7 x 5 2 x 2 )
n
7
2
1
x
Sol.
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FUNCTION
Page # 201
(xviii) If f(x) =
x2
find the domain of
5x
4 & g(x) = x + 3, then
f
g (x).
(iv) f(x) =
Sol.
16
x2
Sol.
(v) f(x) =
1
2 cos 3x
Sol.
3. Find the domain and range of each of the following
functions
(i) f(x) = |x – 3|
Sol.
(ii) f(x) =
1
x
5
Sol.
2
(vi) f(x) = 3 sin
16
x2
Sol.
(iii) f(x) =
1
1
x
Sol.
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FUNCTION
(vii)
Page # 202
sin x
cos x
1 tan2 x
1 cot 2 x
4
2
(x) f(x) = x – 2 x + 5
Sol.
Sol.
(xi) f(x) =
x
1 x2
Sol.
2
4
(viii) f(x) = sin x + cos x
Sol.
(xii) f(x) =
(ix) 3 | sin x | – 4 | cos x |
Sol.
x2
2x
4
2
2x
4
x
Sol.
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FUNCTION
(xiii) f(x) =
Page # 203
2 x +
1
x
(v) f(x) = 3 sin x
Sol.
3
Sol.
(xiv) f(x) =
x 4 3
x 5
(vi) f(x) =
Sol.
x8
x
Sol.
2
(xv) f(x) = log(cosecx – 1) (2 – [sinx] – [sinx] )
Sol.
(vii) f(x) = x + sin x
Sol.
0
4. Make the graph of the following functions
(viii) f(x) = (sinx)
Sol.
(i) f(x) = | x | 3
Sol.
x+5
(ii) f(x) = ln | x |
(ix) f(x) = 3e
Sol.
–7
Sol.
(iii) f(x) = [|x|]
Sol.
(x) f(x) = |sinx| + |cos x|
Sol.
(iv) f(x) = |{x}|
Sol.
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FUNCTION
Page # 204
5. Check whether following pairs of functions are identical or not ?
(i) f(x) =
2
x2 & g(x) = ( x )
Sol.
(ii) f(x) = sec (sec
Sol.
(iii) f(x) =
–1
–1
x) & g(x) = cosec (cosec
(ii) Solve the following equation for
x : 2x + 3[x] – 4 {–x} = 4
(where [ * ] & { * } denotes integral and fractional
part of x)
Sol.
x)
1 cos 2x
& g(x) = cos x
2
Sol.
(iii) The set of real values of ‘x’ satisfying the equality
lnx
(iv) f(x) = x and g(x) = e
Sol.
3
x
4
x
= 5 (where [ * ] denotes the greatest
integer function) belongs to the interval a,
a, b, c
N and
b
where
c
b
is in its lowest form. Find the value
c
of a + b + c + abc.
6. (i) If f(x) =
Sol.
4x
4x
2
, then show that f(x)+f(1 – x)=1
Sol.
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FUNCTION
Page # 205
7. Find whether the following functions are one–one
or many–one
2
(i) f(x) = |x + 5x + 6|
Sol.
(v) f(x) =
1 e
1
1
x
Sol.
(ii) f(x) = | log x |
Sol.
(iii) f(x) = sin 4x, x
,
8 8
Sol.
(iv) f(x) = x +
1
,x
x
(vi) f(x) =
(0,
)
3x 2
– cos x
4
Sol.
Sol.
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FUNCTION
Page # 206
8. Let f : D
R where D is its domain. Find whether
the following functions are into/onto.
(i) f(x) =
1 x6
x3
9. Classify the following functions f(x) defined in
R
R as injective , surjective, both or none.
(i) f(x) = x | x |
Sol.
Sol.
2
(ii) f(x) = x
Sol.
(ii) f(x) = x cos x
Sol.
(iii) f(x) =
x2
1 x2
Sol.
1
(iii) f(x) =
sin | x |
3
2
(iv) f(x) = x – 6 x + 11x – 6
Sol.
Sol.
10. Find fog and gof, if
x
(i) f(x) = e ; g(x) = log x
Sol.
(iv) tan (2 sin x)
Sol.
(ii) f(x) = |x| ; g(x) = sin x
Sol.
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FUNCTION
Page # 207
–1
2
(iii) f(x) = sin x ; g(x) = x
Sol.
(iii) If f(x) = –1 + |x – 2|, 0 x 4
g(x) = 2 – |x|, –1 x 3
Then find fog(x) & gof(x). Draw rough sketch of
the graphs of fog(x) & gof(x).
Sol.
1
2
(iv) f(x) = x + 2 ; g(x) = 1 –
1 x
,x
1
Sol.
11. (i) Let f(x) =
1 x, 0
x
2
3
x
3
x, 2
. Find fof..
Sol.
12. Let f(x) be a polynomial function satisfying the
relation f(x) . f
1
1
= f(x) + f
x
x
x
R – {0} and
f(3) = –26. Determine f (1).
Sol.
1 x if x
(ii) f(x) = x2 if x
0
x if x 1
0 and g(x) = 1 x if x 1
find (fog) (x) and (gof)(x)
Sol.
13. Solve the following problems from (i) to (v) on
functional equation.
(i) The function f(x) defined on the real numbers has
the property that f(f(x)) . (1 + f(x)) = –f(x) for all x in
the domain of f. If the number 3 is the domain and
range of f, compute the value of f(3).
Sol.
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FUNCTION
Page # 208
(ii) Suppose f is a real function satisfying
f(x + f(x)) = 4f(x) and f(1) = 4. Find the value of f(21).
Sol.
(iii) Let ‘f’ be a function defined from
R+
R+. [f(xy)]2 = x(f(y))2 for all positive numbers
x and y and f(2) = 6, find the value of f(50).
Sol.
(v) Let f(x) be function such that f(3) = 1 and
f(3x) = x + f(3x – 3) for all x. Then find the value of
f(300).
Sol.
14. Examine whether the following functions are even
or odd or none.
1 x2
(i) f(x) = log x
Sol.
(ii) f(x) =
(iv) Let f(x) be a function with two properties
(a) for any two real number x and y, f(x + y)=x + f(y)
and (b) f(0) = 2. Find the value of f(100).
Sol.
x(ax
ax
1)
1
Sol.
(iii) f(x) =
x
ex
1
+
x
+1
2
Sol.
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FUNCTION
(iv) f(x) =
Page # 209
Sol.
(1 2x )7
2x
Sol.
(v) f(x) =
sec x x 2
x sin x
9
Sol.
(vi) f(x) =
1 x
x
2
1 x x
2
Sol.
x2 sin x 0 x 1
15. (i) If f(x) =
then extend the
x e x
x 1
definition of f(x) for x
comes
(a) An even function
Sol.
x|x|
,
[1 x] [ x 1] ,
(vii) f(x) =
x|x|
,
x
1
1 x 1
x 1
Sol.
(–
, 0) such that f(x) be-
(b) An odd function
Sol.
(ii) Prove that the function defined as,
(viii) f(x) =
2 x(sin x tan x )
,
x 2
2
3
where [ * ] denotes greatest integer function.
f(x) =
e
|ln{ x }|
{ x}
{ x}
1
|ln{ x }|
where ever it exists
otherwise, then
f(x) is odd as well as even.
(where {x} denotes the fractional part function)
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FUNCTION
Page # 210
Sol.
(iii) f(x) = sin
x
x
+ sin
4
3
Sol.
(iv) f(x) = cos
(iii) Let a and b be real numbers and let
f(x) = a sin x + b3 x + 4,
3
2
x – sin
x.
5
7
Sol.
x R . If f(log10(log310)) = 5
then find the value of f (log10 (log10 3))
Sol.
(v) f(x) = [sin 3x] + |cos 6x|
Sol.
16. Find the period of the following functions
(where [ * ] denotes greatest integer function)
(i) f(x) = 2 + 3 cos (x – 2)
Sol.
(vi) f(x) =
1
1 cos x
Sol.
2
(ii) f(x) = sin 3x + cos x + |tan x|
Sol.
(vii) f(x) =
sin12 x
1 cos2 6x
Sol.
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FUNCTION
Page # 211
2
3
(viii) f(x) = sec x + cosec x
Sol.
(v) f(x) =
1 | sin x |
2 cos x
sin x
| cos x |
Sol.
17. Find the period of the following functions.
(i) f(x) = 1 –
sin2 x
1 cot x
cos 2 x
1 tan x
(vi) f(x) = sin x + tan
Sol.
+ sin
x
2
n 1
+ tan
x
x
x
+ sin 2 + tan 3 + ........
2
2
2
x
2n
Sol.
(ii) f(x) = log (2 + cos 3 x)
Sol.
(iii) f(x) = tan
2
[x],
where [*] denotes greatest integer function
Sol.
(vii) f(x) =
sin x sin 3 x
cos x cos 3x
Sol.
ln sin x
(iv) f(x) e
Sol.
3
+ tan x – cosec (3 x – 5)
18. (a) A function f is defined for all positive integers
and satisfies f(1) = 2005 and
f(1) + f(2) + ...+ f(n) = n2f(n) for all n > 1. Find the
value of f(2004)
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FUNCTION
Page # 212
Sol.
(d) Let P(x) = x6 + ax5 + bx4 + cx 3 + dx2 + ex + f be a
polynomial such that P(1) = 1 ; P(2) = 2; P(3) = 3 ;
P(4) = 4; P(5) = 5 and P(6) = 6 then find the value of
P(7).
Sol.
(b) If a,b are positive real numbers such that a–b= 2,
then find the smallest value of the constant L for
which
x 2 ax
x2
bx < L for all x > 0.
Sol.
19.
Computer the inverse of the function :
(i) f(x) = ln x
x2
1
Sol.
(ii) f(x) =
(c) Let f(x) = x2 + kx ; k is real number. The set of
values of k for which the equation f(x) = 0 and f(f(x)) = 0
have same real solution set.
Sol.
x
2x 1
Sol.
(iii) y =
10x
1
x
1
10
Sol.
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FUNCTION
Page # 213
EXERCISE – IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
1. Which of the following is not a periodic function [AIEEE 2002]
4. The range of the function f(x) =
(A) sin 2x + cos x
(C) tan 4x
Sol.
(A) R
Sol.
(B) cos x
(D) log cos 2x
2. The period of sin2 x is(A) /2
(B)
(C) 3 /2
Sol.
(B) R – {–1}
2 x
,x
2 x
(C) R – {1}
2 is-
[AIEEE-2002]
(D) R – {2}
[AIEEE 2002]
(D) 2
x 2 1 ), is[AIEEE 2003]
(A) neither an even nor an odd function
(B) an even function
(C) an odd function
(D) a periodic function
Sol.
5. The function f(x) = log (x +
3. The function f : R
(A) into
Sol.
(B) onto
R defined by f(x) = sin x is[AIEEE-2002]
(C) one-one
(D) many-one
6. Domain of definition of the function
f(x) =
3
4
x2
(A) (– 1, 0)
(C) ( – 1, 0)
+ log10 (x3 – x), is(1, 2)
(1, 2)
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(2,
)
[AIEEE 2003]
(B) (1, 2)
(D) (1, 2)
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(2,
)
FUNCTION
Page # 214
Sol.
9. The range of the function f(x) = 7– xPx–3 is[AIEEE 2004]
(A) {1, 2, 3}
(B) {1, 2, 3, 4, 5, 6}
(C) {1, 2,3,4}
(D) {1, 2, 3, 4, 5}
Sol.
7. If f : R
R satisfies f(x+ y) = f(x) + f(y), for all x,
n
y
f (r ) is-
R and f(1) = 7, then
[AIEEE 2003]
r 1
(A)
7n (n 1)
2
(B)
7n
2
(C)
7(n 1)
2
(D) 7n (n+1)
Sol.
10.
If f : R
S, defined by f(x) = sin x –
3 cos x+ 1,
is onto, then the interval of S is [AIEEE 2004]
(A) [0, 3]
(B) [–1, 1]
(C) [0, 1]
(D) [–1, 3]
Sol.
8. A function f from the set of natural numbers to
n 1
, when n is odd
2
integers defined by f(n) =
is
n
, when n is even
2
[AIEEE 2003]
(A) neither one-one nor onto
(B) one-one but not onto
(C) onto but not one-one
(D) one-one and onto both
Sol.
11. The graph of the function y = f(x) is symmetrical
about the line x = 2, then[AIEEE 2004]
(A) f(x+ 2) = f(x – 2)
(B) f(2 + x) = f(2 – x)
(C) f(x) = f(–x)
(D) f(x) = – f(–x)
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FUNCTION
Page # 215
Sol.
14. A real valued function f(x) satisfies the functional
equation f(x – y) = f(x) f(y)– f (a–x) f(a + y) where a
is a given constant and f(0)=1, then f(2a – x) is equal
to [AIEEE-2005]
(A) –f(x) (B) f(x) (C) f(a) + f(a – x)
(D) f(–x)
Sol.
sin 1( x 3)
12.
The domain of the function f(x) =
(A) [2,3]
Sol.
(B) [2,3)
(C) [1,2]
9 x2
is-
[AIEEE 2004]
(D) [1, 2)
15.
The largest interval lying in
the function f(x)
13.
Let f : (–1, 1)
f(x) = tan–1
2x
1 x2
B, be a function defined by
, then f is both one-one and onto
when B is the interval (A) 0,
(C)
2
,
2 2
[AIEEE-2005]
(B) 0,
(D)
x
cos 1
2
2
4 x
is defined, is (A) [0,
]
1
,
2
2
for which
log (cos x)
[AIEEE 2007]
(B)
2
,
2
(C)
,
4 2
(D) 0,
Sol.
2
,
2 2
Sol.
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2
FUNCTION
Page # 216
16. L et f : N
Y be a fun ction defin ed as
f(x) = 4x + 3 where Y = |y
x
N : y = 4x + 3 for some
N|. Show that f is invertible and its inverse is
[AIEEE 2008]
(A) g(y) = 4 +
(C) g(y) =
y 3
4
y 3
4
(B) g(y) =
y 3
4
(D) g(y) =
3y 4
3
Sol.
18. Let f(x) = (x + 1)2 –1, x > –1
Statement – 1 :
[AIEEE 2009]
The set {x : f(x) = f–1(x)} = {0, –1}.
Statement – 2 :
f is a bijection.
(A) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1
(B) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.
(C) Statement -1 is true, Statement -2 is false.
(D) Statement -1 is false, Statement -2 is true.
Sol.
1
19.
The domain of the function f(x) =
x
x is :
[AIEEE 2011]
(A) (– , ) (B) (0,
Sol.
17.
) (C) (– , 0) (D) (– ,
) – {0}
For real x, let f(x) = x3 + 5x + 1, then [AIEEE 2009]
(A) f is one – one but not onto R
(B) f is onto R but not one – one
(C) f is one – one and onto R
20.
(D) f is neither one – one nor onto R
Sol.
If a R and the equation
– 3(x – [x])2 + 2(x – [x]) + a2 = 0
(where [x] denotes the greatest integer x)
has no integral solution, then all possible values
of a lie in the interval :
[AIEEE 2014]
(A) (–1, 0)
(0, 1)
(B) (1, 2)
(C) (–2, –1)
(D) (– , –2)
(2, )
Sol.
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FUNCTION
Page # 217
LEVEL – II
JEE ADVANCED
1. If the functions f(x) & g(x) are defined on R
f ( x)
0,
x,
x rational
, g( x)
x irrational
0,
x,
x
x
R such that
irrational
rational
then (f – g) (x) is
[JEE 2005 (Scr.), 1]
(A) one – one and onto (B) neither one–one nor onto
(C) one-one but not onto (D) onto but not one-one
Sol.
3. Let f(x) = x2 and g(x) = sin x for all x R. Then the
set of all x satisfying (f o g o g o f) (x) = (g o g of) (x),
where (f o g) (x) = f(g(x)), is
(A) ± n , n
(C)
2
{0, 1, 2, ...}
+ 2n , n
(D) 2n , n
Sol.
(B) ±
n ,n
{1, 2, ...}
{......, –2, –1, 0, 1, 2 .....}
{......, –2, –1, 0, 1, 2, ....} [JEE 2011]
2. Let S = {1, 2, 3, 4}. The total number of unordered
pairs of disjoint subsets of S is equal to [JEE 2010]
(A) 25
(B) 34
(C) 42
(D) 41
Sol.
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FUNCTION
Page # 218
4. The function f : [0, 3]
[1, 29], defined by
f(x) = 2x3 – 15x2 + 36x + 1, is
[JEE 2012]
(A) one-one and onto. (B) onto but not one-one.
(C) one-one but not onto. (D) neither one-one nor onto
Sol.
5. Let f : (–1, 1)
for
0,
(A) 1 –
(C) 1 –
4
IR be such that f(cos 4 ) =
2
2 sec2
1
,
. Then the value(s) for f
is (are)
4 2
3
3
2
2
3
Sol.
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(B) 1 +
(D) 1 +
3
2
2
3
[JEE 2012]
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FUNCTION
6.
Page # 219
Let f : – 2 , 2
R be given by
7.
f(x) = (log (secx + tanx))3
Then
[JEE 2014]
(A) f(x) is an odd function
(B) f(x) is an one-one function
(C) f(x) is an onto fucntion
(D) f(x) is an even function
Sol.
Let f(x) = sin 6 sin 2 sin x
for all x
R and
g(x) =
sinx for all x R. Let (fog) (x) denote
2
f(g(x)) and (gof) (x) denote g(f(x)). Then
which of the following is (are) true? [JEE 2015]
1 1
(A) Range of f is – ,
2 2
1 1
(B) Range of fog is – 2 , 2
(C) Lim
x 0
f(x)
=
g(x)
6
(D) There is an x
R such that (gof)(x) = 1
Sol.
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FUNCTION
Page # 220
Answer Ex–I
JEE MAIN
1. D
2. D
3. B
4. A
5. A
6. B
7. B
8. D
9. C
10. C
11. A
12. A
13. B
14. B
15. C
16. B
17. D
18. A
19. A
20. C
21. D
22. C
23. C
24. B
25. A
26. D
27. D
28. B
29. D
30. C
31. C
32. C
33. C
Answer Ex–II
JEE ADVANCED (OBJECTIVE)
LEVEL – I
SINGLE CORRECT
1. A
2. A
3. D
4. B
5. D
6. C
7. B
8. C
9. C
10. D
11. B
12. D
13. B
14. C
15. B
16. D
17. C
18. D
19. A
LEVEL – II
MULTIPLE CORRECT
1. A,B
2. B,C,D
3. A,C
8. A,B,C,D
9. A,D
10. A,B,C,D
4. A,C
5. A,B,C
6. A,B,C
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7. B,C,D
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FUNCTION
Page # 221
Answer Ex–III
JEE ADVANCED
Comprehension
1.
C
Matrix Match Type
1.
(A)–S ; (B)–R ; (C)–P ; (D)–Q
2.
(A)–R ; (B)–S ; (C)–P ; (D)–Q
1.
(i) R – {–1, 1}
(ii) (0, )
(iii) R
(iv) [–2, 0)
(vi) [0, 1]
(vii) [–1, 1]
(viii) R
(ix)
[4, 5)
(xiii) (2, 3)
Subjective Type
1
,1
2
(v)
1,
3
2
n ,n
(x)
2.
5
3
,
4
4
(i)
,
4 4
(iv) (– , – 1)
(vi) 0,
1
100
1
(viii)
2
(x) {4}
(xii)
[0,
)
1
,0
[5,
1
,
6 3
3 5
,
4 4
3.
C
I (xii) (0, 1)
(ii)
4,
1
2
(2,
(v) (3 2 < x < 3
1
1
,
100 10
5
D
(xi) R – {2n }, n
4
n I
2.
5
2
)
(iii) (–
) U (3 < x
(xi) (0 , 1/4) U (3/4 , 1) U {x : x
5
,6
3
N, x
2}
(xiii)
but x 1 where K is non negative integer
(xv) {x 1000
x < 10000}
(xvii) (1, 2)
2,
5
2
4)
(ix) ( 3, 1] U {0} U [1,3)
)
(xiv) 2K < x < (2K + 1)
, – 3]
(vii) ( 1 < x < 1/2) U (x > 1)
,
(xvi) (–2, –1)
(xviii) (
(–1, 0)
, 3)
(1, 2)
( 3 , 1]
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[4 ,
(0, 1)
)
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FUNCTION
3.
Page # 222
(i) R, [0,
)
(ii) (5,
(iv) [–4, 4], [0, 4]
n
(vii) R– 2 , n
(x) R, [4,
), (0, )
(iii) [0,
1
,1
3
(v) R,
(vi)
3
z , [–1, 1] (viii) R,
,1
4
)
(xiii) [–1, 2], [ 3 ,
6]
(xv) (2n , (2n + 1) ) – {2n +
(ii) Y
2
1
,3
3
(xii) R,
(xiv) [–4,
1
) – {5}, 0, 6
, 2n +
(iii) N
6
, 2n +
5
}, R – {0}
6
(i) N
7.
(i) many–one (ii) many–one
(iii) one–one (iv) many–one
8.
(i) into
(iii) into
9.
(i) one-one, onto
10.
(i) fog = x, x > 0 ; gof = x, x
–1
(iii) sin
2
(x ), (sin
(iv) N
1 1
,
6 3
5.
(ii) onto
(ii) many-one, into
–1
x)
2
R
3
,
, 0,
4 4
2
(ix) R, [–4, 3]
1 1
,
2 2
(xi) R,
), (0, 1]
6.
(ii)
3
2
(iii) 20
(v) one–one (vi) many–one
(iv) onto
(iii) Many-one, into
(iv) Many-one, onto
(ii) |sin x|, sin |x|
(iv)
3x 2 4x 2 x2
, 2
(1 x )2
x
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2
1
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FUNCTION
Page # 223
2
11.
x ,
(i) (fof) (x) = 2
4
0
x
1
x , 1 x
2
x , 2
3
x
x
if
x 0
x2 if
x 0
2
x
if
0
x
1
1
x if 0 x 1
(ii) (gof) (x) =
,
(fog)
(x)
=
x
if
x 1
1 x2 if
x 1
x
3
1 x 0
0 x 2 , (gof) (x) = x
5
(1 x),
x 1,
(iii) (fog) (x) =
–3
13.
14.
(i) odd
(ii) even
(iii) even
(vi) odd
(vii) even
(viii) odd
x2
(i)
3
4
12.
sin x
x ex
15.
(i) (a) f(x) =
16.
(i) 2
(ii) 2
(vii) /12
(viii) 2
17.
(i)
(ii)
18.
(a)
19.
(i)
1
1002
ex
e
2
2
3
(ii) 64
x
x 1
x 2
x 3
x 4
x2 sin x
x ex
(iv) 70
(iii) 2
(iv) 2
(c) [0, 4)
(iii)
(iv) 102
(v) 5050
(iv) neither even nor odd
(iii) 24
log2 x
(ii) log x 1
2
0
1
2
3
(iii) 30
1 x 0
(b) f(x) =
x
1
(b) 1
1,
x,
1,
x,
1 x 0
x
1
(v)
(v) 2
(v) even
(iii) 3
2
3
(vi) 2
n
(vi) 2
(vii)
(d) 727
1
1 x
log
2
1 x
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FUNCTION
Page # 224
Answer Ex–IV
PREVIOUS YEARS
LEVEL – I
JEE MAIN
1. B
2. B
3. A,D
4. B
5. C
6. A
7. A
8. D
9. A
10. D
11. B
12. B
13. D
14. A
15. D
16. C
17. C
18. B
19. C
20. A
LEVEL – II
1.
7.
A
2.
A, B, C
D
JEE ADVANCED
3.
A
4.
B
5.
A,B
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6.
A,B,C
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