MOTION SAMPLE BOOKLET [CLASS XII] • GRAVITATION • CHEMICAL KINETICS • FUNCTION Copyright © reserved with Motion Edu. Pvt. Ltd. and Publications All rights reserved. No part of this work herein should be reproduced or used either graphically, electronically, mechanically or by recording, photocopying, taping, web distributing or by storing in any form and retrieving without the prior written permission of the publisher. Anybody violating this is liable to be legally prosecuted. Cor porate Head Office 394 - Raj eev G and hi Nagar K o ta, (R aj.) Ph. No. : 08003899588, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motio niitjee.c om , [email protected] niitjee.co m 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, Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 6 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 8 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 ) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 9 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 10 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION 5. 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] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, ...(2) : [email protected] GRAVITATION Page # 12 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 14 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 15 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 16 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) m2 GRAVITATION Page # 17 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, GMe then its path will be elliptical r : [email protected] GRAVITATION Page # 18 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 19 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 20 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 21 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 22 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 23 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 24 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 26 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, so T2 r7/2 : [email protected] GRAVITATION Page # 28 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 29 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 30 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 31 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 32 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 33 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 34 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 35 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 36 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 37 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 38 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) GRAVITATION Page # 39 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 48 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) (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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 54 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] GRAVITATION Page # 56 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, : [email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 62 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 64 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 66 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 : Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 68 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 69 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 70 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 71 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 72 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 73 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 74 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 ) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 76 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 77 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 78 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 79 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 80 CHEMICAL KINETICS (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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 81 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, B) remains close to equilibrium. :[email protected] Page # 82 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 83 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 84 CHEMICAL KINETICS 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 85 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 86 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 87 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 88 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 90 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 91 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 92 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 94 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 96 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 98 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 100 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.) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 102 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 104 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, –3 –3 3.2 × 10 –2 10 –3 :[email protected] –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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 112 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 114 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 116 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 117 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 118 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 120 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 122 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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- : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 126 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] T3 /4 hr Page # 128 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 130 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 132 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS 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? : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 134 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 136 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 138 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) CHEMICAL KINETICS Page # 139 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 140 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] Page # 144 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, Domain = ] 4, 5 [ U ] 6, :[email protected] [ 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, O (1, 0) :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, –5/4 –1 t 0 1 :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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). : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, x = – x. 1 x 2 2 2 :[email protected] 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) Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 . : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) 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 . : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] ...(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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, x2 3x+y=x :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] ) 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, x is 2 1 { 2 (sin x –cos x) + 3} is (A) [0, 1] (B) [0, 2] (C) 0, cos :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 4 x is (C) –1 (D) x 2 :[email protected] 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, ) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 2 (D) x – 2 :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (C) (0,2 2 ] (D) (0, 2 3 ] :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 = : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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, ¥) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (2, ) [AIEEE 2003] (B) (1, 2) (D) (1, 2) :[email protected] (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) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, (B) 1 + (D) 1 + 3 2 2 3 [JEE 2012] :[email protected] 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. : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 7. B,C,D :[email protected] 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] : 0744-2209671, 08003899588 | url : www.motioniitjee.com, [4 , (0, 1) ) :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 2 1 :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected] 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 : 0744-2209671, 08003899588 | url : www.motioniitjee.com, 6. A,B,C :[email protected]

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