2.2 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words, ∆ x v = lim (2.1a) ∆ t → 0 ∆ t Fig. 2.1 Determining velocity from position-time d x = (2.1b) graph. Velocity at t = 4 s is the slope of the dt tangent to the graph at that instant. lim where the symbol stands for the operation ∆→t 0 Now, we decrease the value of ∆t from 2 s to 1 of taking limit as ∆tg0 of the quantity on its s. Then line P1P2 becomes Q1Q2 and its slope right. In the language of calculus, the quantity gives the value of the average velocity over on the right hand side of Eq. (2.1a) is the the interval 3.5 s to 4.5 s. In the limit ∆t → 0, differential coefficient of x with respect to t and the line P1P2 becomes tangent to the position- d x time curve at the point P and the velocity at t is denoted by (see Appendix 2.1). It is the d t = 4 s is given by the slope of the tangent at rate of change of position with respect to time, that point. It is difficult to show this process graphically. But if we useat that instant. numerical method to obtain the value of We can use Eq. (2.1a) for obtaining the the velocity, the meaning of the limiting value of velocity at an instant either process becomes clear. For the graph shown graphically or numerically. Suppose that we in Fig. 2.1, x = 0.08 t3. Table 2.1 gives the want to obtain graphically the value of value of ∆x/∆t calculated for ∆t equal to 2.0 s, velocity at time t = 4 s (point P) for the motion 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = of the car represented in Fig.2.1 calculation. 4.0 s. The second and third columns give theLet us take ∆t = 2 s centred at t = 4 s. Then, t  t    ∆ ∆by the definition of the average velocity, the t + t −     and t 2 = and the value of t1= 2 2slope of ( Fig. 2.1) gives the value of     line P1P2 average velocity over the interval 3 s to 5 s. fourth and the fifth columns give the ∆x Table 2.1 Limiting value of at t = 4 s ∆ t Reprint 2025-26 MOTION IN A STRAIGHT LINE 15 3 a + 16b – a – 4b corresponding values of x, i.e. x (t1) = 0.08 t1 = = 6.0 × b 2.0 -1 ⊳and x (t2) = 0.08 t23. The sixth column lists the = 6.0 × 2.5 =15 m s difference ∆x = x (t2) – x (t1) and the last column gives the ratio of ∆x and ∆t, i.e. the Note that for uniform motion, velocity is average velocity corresponding to the value the same as the average velocity at all of ∆t listed in the first column. instants. We see from Table 2.1 that as we decrease Instantaneous speed or simply speed is the the value of ∆t from 2.0 s to 0.010 s, the value of magnitude of velocity. For example, a velocity of the average velocity approaches the limiting + 24.0 m s–1 and a velocity of – 24.0 m s–1 — value 3.84 m s–1 which is the value of velocity at both have an associated speed of 24.0 m s-1. It should be noted that though average speed over dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this a finite interval of time is greater or equal to the dt magnitude of the average velocity, manner, we can calculate velocity at each instantaneous speed at an instant is equal to instant for motion of the car. the magnitude of the instantaneous velocity at The graphical method for the determination that instant. Why so ? of the instantaneous velocity is always not a 2.3 ACCELERATIONconvenient method. For this, we must carefully plot the position–time graph and calculate the The velocity of an object, in general, changes value of average velocity as ∆t becomes smaller during its course of motion. How to describe and smaller. It is easier to calculate the value this change? Should it be described as the rate of velocity at different instants if we have data of change in velocity with distance or with of positions at different instants or exact time ? This was a problem even in Galileo’s expression for the position as a function of time. time. It was first thought that this change could Then, we calculate ∆x/∆t from the data for be described by the rate of change of velocity decreasing the value of ∆t and find the limiting with distance. But, through his studies of value as we have done in Table 2.1 or use motion of freely falling objects and motion of differential calculus for the given expression and objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is dx calculate at different instants as done in a constant of motion for all objects in free fall. dt On the other hand, the change in velocity with the following example. distance is not constant – it decreases with the ⊳ increasing distance of fall. This led to the Example 2.1 The position of an object concept of acceleration as the rate of change moving along x-axis is given by x = a + bt2 of velocity with time. where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time interval measured in seconds. What is its velocity at is defined as the change of velocity divided by t = 0 s and t = 2.0 s. What is the average the time interval : velocity between t = 2.0 s and t = 4.0 s ? v 2 – v1 ∆v (2.2)Answer In notation of differential calculus, the a = = t 2 – t1 ∆tvelocity is where v2 and v1 are the instantaneous velocities dx d 2 -1 2b t = 5.0 t m s or simply velocities at time t2 and t1 . It is thev = = ( a + bt ) = dt dt average change of velocity per unit time. The SI At t = 0 s, v = 0 m s–1 and at t = 2.0 s, unit of acceleration is m s–2 . v = 10 m s-1 . On a plot of velocity versus time, the average acceleration is the slope of the straight line x ( 4.0 ) − x ( 2.0 )Average velocity = connecting the points corresponding to (v2, t2) 4.0 − 2.0 and (v1, t1). Reprint 2025-26 16 PHYSICS Instantaneous acceleration is defined in the (c) An object is moving in negative direction same way as the instantaneous velocity : with a negative acceleration. ∆v d v (d) An object is moving in positive direction lim a = = (2.3) till time t1, and then turns back with the d t ∆→ t 0 ∆ t same negative acceleration. The acceleration at an instant is the slope of the tangent to the v–t curve at that An interesting feature of a velocity-time instant. graph for any moving object is that the area Since velocity is a quantity having both under the curve represents the magnitude and direction, a change in displacement over a given time interval. A velocity may involve either or both of these general proof of this statement requires use of factors. Acceleration, therefore, may result calculus. We can, however, see that it is true from a change in speed (magnitude), a for the simple case of an object moving with change in direction or changes in both. Like constant velocity u. Its velocity-time graph is velocity, acceleration can also be positive, as shown in Fig. 2.4. negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is vo at t = 0 and v at time t, we have v − v 0 a = or, v = v 0 + a t (2.4) t − 0 Fig. 2.3 Velocity–time graph for motions with Fig. 2.2 Position-time graph for motion with constant acceleration. (a) Motion in positive (a) positive acceleration; (b) negative direction with positive acceleration, acceleration, and (c) zero acceleration. (b) Motion in positive direction with Let us see how velocity-time graph looks like negative acceleration, (c) Motion in for some simple cases. Fig. 2.3 shows velocity- negative direction with negative acceleration, (d) Motion of an object withtime graph for motion with constant acceleration negative acceleration that changesfor the following cases : direction at time t1. Between times 0 to (a) An object is moving in a positive direction t1, it moves in positive x - direction with a positive acceleration. and between t1 and t2 it moves in the (b) An object is moving in positive direction opposite direction. with a negative acceleration. Reprint 2025-26 MOTION IN A STRAIGHT LINE 17 Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which Fig. 2.5 Area under v-t curve for an object with is the displacement in this time interval. How uniform acceleration. come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at As explained in the previous section, the area the answer. under v-t curve represents the displacement. Therefore, the displacement x of the object is : Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp 1 x = ( v – v 0 ) t + v 0 t (2.5)kinks at some points implying that the 2 functions are not differentiable at these But v − v 0 = a tpoints. In any realistic situation, the functions will be differentiable at all points 1 2 Therefore, x = a t + v 0 tand the graphs will be smooth. 2 What this means physically is that 1 2 or, x = v 0 t + at (2.6)acceleration and velocity cannot change 2 values abruptly at an instant. Changes are Equation (2.5) can also be written as always continuous. v + v 0 x = t = v t (2.7a)2.4 KINEMATIC EQUATIONS FOR 2 UNIFORMLY ACCELERATED MOTION where, For uniformly accelerated motion, we can derive some simple equations that relate displacement v + v 0 v = (constant acceleration only)(x), time taken (t), initial velocity (v0), final 2 velocity (v) and acceleration (a). Equation (2.4) (2.7b) already obtained gives a relation between final and initial velocities v and v0 of an object moving Equations (2.7a) and (2.7b) mean that the object with uniform acceleration a : has undergone displacement x with an average velocity equal to the arithmetic average of the v = v0 + at (2.4) initial and final velocities. From Eq. (2.4), t = (v – v0)/a. Substituting this in This relation is graphically represented in Fig. 2.5. Eq. (2.7a), we get The area under this curve is : Area between instants 0 and t = Area of triangle  v + v 0   v − v 0  v 2 − v 02 x = v t =ABC + Area of rectangle OACD  2   a = 2a 1 2 2 = (v – v 0 ) t + v 0 t v = v 0 + 2ax (2.8) 2 Reprint 2025-26 18 PHYSICS This equation can also be obtained by t v 0 + at ) d tsubstituting the value of t from Eq. (2.4) into Eq. = ∫ 0 ( (2.6). Thus, we have obtained three important equations : 1 2 x – x 0 = v 0 t + a t 2 v = v 0 + at 1 2 1 2 x = x 0 + v 0 t + a t x = v 0t + at 2 2 We can write v 2 = v 02 + 2ax (2.9a) d v d v d x d v a = = = v d t d x d t d x connecting five quantities v0, v, a, t and x. These or, v dv = a dxare kinematic equations of rectilinear motion for Integrating both sides,constant acceleration. The set of Eq. (2.9a) were obtained by v x v d v = a d xassuming that at t = 0, the position of the particle, ∫ v 0 ∫ x 0 x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non- v 2 – v 02 = a ( x – x 0 ) zero, say x0. Then Eqs. (2.9a) are modified 2 (replacing x by x – x0 ) to : 2 2 v = v 0 + 2a ( x – x 0 ) v = v 0 + at The advantage of this method is that it can be used 1 2 for motion with non-uniform acceleration x = x 0 + v 0t + at (2.9b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (2.9c) important cases. ⊳ ⊳ ⊳ Example 2.3 A ball is thrown vertically Example 2.2 Obtain equations of motion upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how longAnswer By definition will it be before the ball hits the ground? d v Take g = 10 m s–2. a = d t dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the v t ∫ v 0 d v = ∫ 0 a d t ground, as shown in Fig. 2.6. Now vo = + 20 m s–1, t d t (a is a = – g = –10 m s–2, = a ∫ 0 v = 0 m s–1 constant) If the ball rises to height y from the point of v – v 0 = at launch, then using the equation 2 + 2 a 0 ( y – y 0 ) v = v 0 + at v 2 = v we get d x Further, v = 0 = (20)2 + 2(–10)(y – y0) d t Solving, we get, (y – y0) = 20 m. dx = v dt Integrating both sides (b) We can solve this part of the problem in two x t ways. Note carefully the methods used. ∫ x 0 dx = ∫v0 d t Reprint 2025-26 MOTION IN A STRAIGHT LINE 19 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. ⊳ ⊳ Example 2.4 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 2.6 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use v = 0 – g t = –9.8 t m s–1equation y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y 0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t2 2 the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, with time have been plotted in Fig. 2.7(a), (b)Therefore, the total time taken by the ball before and (c). it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y = 0 m vo = 20 m s-1, a = –10m s–2, t = ? (a) Reprint 2025-26 20 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 2 y = −1 gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ… which are given in second column of Table 2.2. If we take (–1/ 2) gτ2 as y0 — the position coordinate after first time interval τ, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. ⊳ ⊳ Example 2.6 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and theFig. 2.7 Motion of an object under free fall. braking capacity, or deceleration, –a that (a) Variation of acceleration with time. (b) Variation of velocity with time. is caused by the braking. Derive an (c) Variation of distance with time ⊳ expression for stopping distance of a vehicle in terms of vo and a. ⊳ Example 2.5 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling 2 motion v2 = vo + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it. – v 02 d s = 2aAnswer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals τ and find out the distances the square of the initial velocity. Doubling the Table 2.2 Reprint 2025-26 MOTION IN A STRAIGHT LINE 21 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example, in school zones. ⊳ ⊳ Example 2.7 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think Fig. 2.8 Measuring the reaction time. and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before Answer The ruler drops under free fall. he slams the brakes of the car is the Therefore, vo = 0, and a = – g = –9.8 m s–2. The reaction time. Reaction time depends distance travelled d and the reaction time tr are on complexity of the situation and on related by an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between Or, your thumb and forefinger (Fig. 2.8). Given d = 21.0 cm and g = 9.8 m s–2 the reaction After you catch it, find the distance d time is travelled by the ruler. In a particular case, d was found to be 21.0 cm. ⊳ Estimate reaction time. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small : ∆ x d x v = lim v = lim = ∆→t 0 ∆→t 0 ∆t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. Reprint 2025-26 22 PHYSICS 3. Average acceleration is the change in velocity divided by the time interval during which the change occurs : ∆ v a = ∆t 4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆v d v a = lim a = lim = ∆→t 0 ∆→t 0 ∆ t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 5. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at 1 2 x = v0 t + at 2 v 2 = v 02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x – x0). Reprint 2025-26 MOTION IN A STRAIGHT LINE 23 POINTS TO PONDER 1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. Reprint 2025-26 24 PHYSICS EXERCISES 2.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 2.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice). Fig. 2.9 2.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 2.5 A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? 2.6 A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). 2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. Reprint 2025-26 MOTION IN A STRAIGHT LINE 25 2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. 2.9 Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. 2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] Fig. 2.10 2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. 2.13 Figure 2.11shows the x-t plot of one- dimensional motion of a particle. Is it correct to say from the graph that the particle moves Fig. 2.11 in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 2.14 A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car). Reprint 2025-26 26 PHYSICS 2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12): Fig. 2.12 2.16 Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s. Fig. 2.13 2.17 Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 2.14 2.18 Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Fig. 2.15 Reprint 2025-26 CHAPTER THREE MOTION IN A PLANE 3.1 INTRODUCTION In the last chapter we developed the concepts of position, displacement, velocity and acceleration that are needed to 3.1 Introduction describe the motion of an object along a straight line. We 3.2 Scalars and vectors found that the directional aspect of these quantities can be taken care of by + and – signs, as in one dimension only two3.3 Multiplication of vectors by real numbers directions are possible. But in order to describe motion of an 3.4 Addition and subtraction of object in two dimensions (a plane) or three dimensions vectors — graphical method (space), we need to use vectors to describe the above- 3.5 Resolution of vectors mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors. What is a vector? How to3.6 Vector addition — analytical method add, subtract and multiply vectors ? What is the result of 3.7 Motion in a plane multiplying a vector by a real number ? We shall learn this to enable us to use vectors for defining velocity and3.8 Motion in a plane with constant acceleration acceleration in a plane. We then discuss motion of an object 3.9 Projectile motion in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail3.10 Uniform circular motion the projectile motion. Circular motion is a familiar class of Summary motion that has a special significance in daily-life situations. Points to ponder We shall discuss uniform circular motion in some detail. Exercises The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions. 3.2 SCALARS AND VECTORS In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar. A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided Reprint 2025-26 28 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP′ if the length and breadth of a rectangle are denoted by r′. The length of the vector r 1.0 m and 0.5 m respectively, then its represents the magnitude of the vector and its perimeter is the sum of the lengths of the direction is the direction in which P lies as seen four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = from O. If the object moves from P to P′, the 3.0 m. The length of each side is a scalar vector PP′ (with tail at P and tip at P′) is called and the perimeter is also a scalar. Take the displacement vector corresponding to another example: the maximum and motion from point P (at time t) to point P′ (at time t′). minimum temperatures on a particular day are 35.6 °C and 24.2 °C respectively. Then, the difference between the two temperatures is 11.4 °C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of

3.10 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below: A/ mol L–1 0.20 0.20 0.40 B/ mol L–1 0.30 0.10 0.05 r0/mol L–1s–1 5.07 × 10–5 5.07 × 10–5 1.43 × 10–4 What is the order of the reaction with respect to A and B? 3.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B ® C + D Experiment [A]/mol L–1 [B]/mol L–1 Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 × 10–3 II 0.3 0.2 7.2 × 10–2 III 0.3 0.4 2.88 × 10–1 IV 0.4 0.1 2.40 × 10–2 Determine the rate law and the rate constant for the reaction. 3.12 The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/ mol L–1 [B]/ mol L–1 Initial rate/ mol L–1 min–1 I 0.1 0.1 2.0 × 10–2 II – 0.2 4.0 × 10–2 III 0.4 0.4 – IV – 0.2 2.0 × 10–2 3.13 Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s–1 (ii) 2 min–1 (iii) 4 years–1 3.14 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. 3.15 The experimental data for decomposition of N2O5 [2N2O5 ® 4NO2 + O2] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 102 × [N2O5]/ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 mol L–1 (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N2O5] and t. (iv) What is the rate law ? Chemistry 86 Reprint 2025-26 (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

2.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × k/S m–1 1.237 11.85 23.15 55.53 106.74 Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0m . 2.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant? 2.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al? (ii) 1 mol of Cu2+ to Cu? (iii) 1 mol of MnO4– to Mn2+? 2.13 How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl2? (ii) 40.0 g of Al from molten Al2O3? 2.14 How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to O2? (ii) 1 mol of FeO to Fe2O3? 2.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? 2.16 Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? 2.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I–(aq) (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) (iv) Ag(s) and Fe 3+ (aq) (v) Br2 (aq) and Fe2+ (aq). 2.18 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. Answers to Some Intext Questions 2.5 E(cell) = 0.91 V 2.6 ∆ rG o = −45.54 kJ mol −1 , Kc = 9.62 ×107 2.9 0.114, 3.67 × 10–4 mol L–1 Chemistry 60 Reprint 2025-26 UnitUnitUnit Unit33Unit Objectives ChemicalChemical KineticsKinetics After studying this Unit, you will be able to · define the average and Chemical Kinetics helps us to understand how chemical reactions instantaneous rate of a reaction; occur. · express the rate of a reaction in terms of change in concentration Chemistry, by its very nature, is concerned with change. of either of the reactants or Substances with well defined properties are converted products with time; by chemical reactions into other substances with · distinguish between elementary different properties. For any chemical reaction, chemists and complex reactions; try to find out · differentiate between the (a) the feasibility of a chemical reaction which can be molecularity and order of a reaction; predicted by thermodynamics ( as you know that a · define rate constant; reaction with DG < 0, at constant temperature and pressure is feasible);· discuss the dependence of rate of reactions on concentration, (b) extent to which a reaction will proceed can be temperature and catalyst; determined from chemical equilibrium; · derive integrated rate equations (c) speed of a reaction i.e. time taken by a reaction to for the zero and first order reach equilibrium. reactions; Along with feasibility and extent, it is equally · determine the rate constants for important to know the rate and the factors controlling zeroth and first order reactions; the rate of a chemical reaction for its complete · describe collision theory. understanding. For example, which parameters determine as to how rapidly food gets spoiled? How to design a rapidly setting material for dental filling? Or what controls the rate at which fuel burns in an auto engine? All these questions can be answered by the branch of chemistry, which deals with the study of reaction rates and their mechanisms, called chemical kinetics. The word kinetics is derived from the Greek word ‘kinesis’ meaning movement. Thermodynamics tells only about the feasibility of a reaction whereas chemical kinetics tells about the rate of a reaction. For example, thermodynamic data indicate that diamond shall convert to graphite but in reality the conversion rate is so slow that the change is not perceptible at all. Therefore, most people think Reprint 2025-26 that diamond is forever. Kinetic studies not only help us to determine the speed or rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction. At the macroscopic level, we are interested in amounts reacted or formed and the rates of their consumption or formation. At the molecular level, the reaction mechanisms involving orientation and energy of molecules undergoing collisions, are discussed. In this Unit, we shall be dealing with average and instantaneous rate of reaction and the factors affecting these. Some elementary ideas about the collision theory of reaction rates are also given. However, in order to understand all these, let us first learn about the reaction rate. 3.13.13.13.13.1 RateRateRateRateRate ofofofofof aaaaa Some reactions such as ionic reactions occur very fast, for example, ChemicalChemicalChemicalChemicalChemical precipitation of silver chloride occurs instantaneously by mixing of aqueous solutions of silver nitrate and sodium chloride. On the other ReactionReactionReactionReactionReaction hand, some reactions are very slow, for example, rusting of iron in the presence of air and moisture. Also there are reactions like inversion of cane sugar and hydrolysis of starch, which proceed with a moderate speed. Can you think of more examples from each category? You must be knowing that speed of an automobile is expressed in terms of change in the position or distance covered by it in a certain period of time. Similarly, the speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R ® P One mole of the reactant R produces one mole of the product P. If [R]1 and [P]1 are the concentrations of R and P respectively at time t1 and [R]2 and [P]2 are their concentrations at time t2 then, Dt = t2 – t1 D[R] = [R]2 – [R]1 D [P] = [P]2 – [P]1 The square brackets in the above expressions are used to express molar concentration. Rate of disappearance of R Decrease in concentration of R ∆ [ R ] = = − (3.1) Time taken ∆ t Chemistry 62 Reprint 2025-26 Rate of appearance of P Increase in concentration of P ∆ [ P ] = = + (3.2) Time taken ∆t Since, D[R] is a negative quantity (as concentration of reactants is decreasing), it is multiplied with –1 to make the rate of the reaction a positive quantity. Equations (3.1) and (3.2) given above represent the average rate of a reaction, rav. Average rate depends upon the change in concentration of reactants or products and the time taken for that change to occur (Fig. 3.1). { } Fig. 3.1: Instantaneous and average rate of a reaction Units of rate of a reaction From equations (3.1) and (3.2), it is clear that units of rate are concentration time–1. For example, if concentration is in mol L–1 and time is in seconds then the units will be mol L-1s–1. However, in gaseous reactions, when the concentration of gases is expressed in terms of their partial pressures, then the units of the rate equation will be atm s–1. From the concentrations of C4H9Cl (butyl chloride) at different times given ExampleExampleExampleExampleExample 3.13.13.13.13.1 below, calculate the average rate of the reaction: C4H9Cl + H2O ® C4H9OH + HCl during different intervals of time. t/s 0 50 100 150 200 300 400 700 800 [C4H9Cl]/mol L–1 0.100 0.0905 0.0820 0.0741 0.0671 0.0549 0.0439 0.0210 0.017 We can determine the difference in concentration over different intervals SolutionSolutionSolutionSolutionSolution of time and thus determine the average rate by dividing D[R] by Dt (Table 3.1). 63 Chemical Kinetics Reprint 2025-26 Table 3.1: Average rates of hydrolysis of butyl chloride [C4H9CI]t1 / [C4H9CI]t2 / t1/s t2/s rav × 104/mol L–1s–1 × 10 4 mol L–1 mol L–1 = – / ( t 2 − t1 ) C 4 H 9 Cl ]t 2 – [ C 4 H 9 Cl ]t1 {[ } 0.100 0.0905 0 50 1.90 0.0905 0.0820 50 100 1.70 0.0820 0.0741 100 150 1.58 0.0741 0.0671 150 200 1.40 0.0671 0.0549 200 300 1.22 0.0549 0.0439 300 400 1.10 0.0439 0.0335 400 500 1.04 0.0210 0.017 700 800 0.4 It can be seen (Table 3.1) that the average rate falls from 1.90 × 0-4 mol L-1s-1 to 0.4 × 10-4 mol L-1s-1. However, average rate cannot be used to predict the rate of a reaction at a particular instant as it would be constant for the time interval for which it is calculated. So, to express the rate at a particular moment of time we determine the instantaneous rate. It is obtained when we consider the average rate at the smallest time interval say dt ( i.e. when Dt approaches zero). Hence, mathematically for an infinitesimally small dt instantaneous rate is given by −∆ [ R ] ∆ [ P ] rav = = (3.3) ∆t ∆ t  d  R  d P As Dt ® 0 or rinst   d t d t Fig 3.2 Instantaneous rate of hydrolysis of butyl chloride(C4H9Cl) Chemistry 64 Reprint 2025-26 It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 3.1). So in problem 3.1, rinst at 600s for example, can be calculated by plotting concentration of butyl chloride as a function of time. A tangent is drawn that touches the curve at t = 600 s (Fig. 3.2). The slope of this tangent gives the instantaneous rate. So, rinst at 600 s = – mol L–1 = 5.12 × 10–5 mol L–1s–1 At t = 250 s rinst = 1.22 × 10–4 mol L–1s–1 t = 350 s rinst = 1.0 × 10–4 mol L–1s–1 t = 450 s rinst = 6.4 ×× 10–5 mol L–1s–1 Now consider a reaction Hg(l) + Cl2 (g) ® HgCl2(s) Where stoichiometric coefficients of the reactants and products are same, then rate of the reaction is given as ∆ [ Hg ] ∆ [ Cl 2 ] ∆ [ HgCl 2 ] Rate of reaction = – = – = ∆t ∆t ∆ t i.e., rate of disappearance of any of the reactants is same as the rate of appearance of the products. But in the following reaction, two moles of HI decompose to produce one mole each of H2 and I2, 2HI(g) ® H2(g) + I2(g) For expressing the rate of such a reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or the rate of appearance of products is divided by their respective stoichiometric coefficients. Since rate of consumption of HI is twice the rate of formation of H2 or I2, to make them equal, the term D[HI] is divided by 2. The rate of this reaction is given by 1 ∆ [ HI ] ∆ [ H 2 ] ∆ [ I 2 ] Rate of reaction = − = = 2 ∆ t ∆ t ∆ t Similarly, for the reaction 5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) ® 3 Br2 (aq) + 3 H2O (l) − − + 1 ∆ [ Br BrO 3 1 ∆ [ H ] 1 ∆ [ Br2 ] 1 ∆ [ H 2 O ] ] ∆  Rate = − = − = − = = 5 ∆ t ∆ t 6 ∆t 3 ∆ t 3 ∆t For a gaseous reaction at constant temperature, concentration is directly proportional to the partial pressure of a species and hence, rate can also be expressed as rate of change in partial pressure of the reactant or the product. 65 Chemical Kinetics Reprint 2025-26 ExampleExampleExampleExampleExample 3.23.23.23.23.2 The decomposition of N2O5 in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2.33 mol L–1 and after 184 minutes, it is reduced to 2.08 mol L–1. The reaction takes place according to the equation 2 N2O5 (g) ® 4 NO2 (g) + O2 (g) Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period? 1  ( 2.08 − 2 .33 ) mol L−1  1 ∆ [ N 2 O5 ] SolutionSolutionSolutionSolutionSolution Average Rate = − = − 184 min 2  ∆t  2   = 6.79 × 10–4 mol L–1/min = (6.79 × 10–4 mol L–1 min–1) × (60 min/1h) = 4.07 × 10–2 mol L–1/h = 6.79 × 10–4 mol L–1 × 1min/60s = 1.13 × 10–5 mol L–1s–1 It may be remembered that 1 ∆ [ NO 2 ] Rate = 4  ∆t  ∆ [ NO 2 ] = 6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1 ∆t IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.1 For the reaction R ® P, the concentration of a reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. 3.2 In a reaction, 2A ® Products, the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval? 3.23.23.23.23.2 FactorsFactorsFactorsFactorsFactors InfluencingInfluencingInfluencingInfluencingInfluencing Rate of reaction depends upon the experimental conditions such RateRateRateRateRate ofofofofof aaaaa ReactionReactionReactionReactionReaction as concentration of reactants (pressure in case of gases), temperature and catalyst. 3.2.1 Dependence The rate of a chemical reaction at a given temperature may depend on of Rate on the concentration of one or more reactants and products. The Concentration representation of rate of reaction in terms of concentration of the reactants is known as rate law. It is also called as rate equation or rate expression. 3.2.2 Rate The results in Table 3.1 clearly show that rate of a reaction decreases with Expression the passage of time as the concentration of reactants decrease. Conversely, and Rate rates generally increase when reactant concentrations increase. So, rate of Constant a reaction depends upon the concentration of reactants. Chemistry 66 Reprint 2025-26 Consider a general reaction aA + bB ® cC + dD where a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is Rate µ [A] x [B] y (3.4) where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Above equation can also be written as Rate = k [A] x [B] y (3.4a) d [ R ] x y − = k [ A ] [ B ] (3.4b) d t This form of equation (3.4 b) is known as differential rate equation, where k is a proportionality constant called rate constant. The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression. Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. For example: 2NO(g) + O2(g) ® 2NO2 (g) We can measure the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. The following results are obtained (Table 3.2). Table 3.2: Initial rate of formation of NO2 Experiment Initial [NO]/ mol L-1 Initial [O2]/ mol L-1 Initial rate of formation of NO2/ mol L-1s-1 1. 0.30 0.30 0.096 2. 0.60 0.30 0.384 3. 0.30 0.60 0.192 4. 0.60 0.60 0.768 It is obvious, after looking at the results, that when the concentration of NO is doubled and that of O2 is kept constant then the initial rate increases by a factor of four from 0.096 to 0.384 mol L–1s–1. This indicates that the rate depends upon the square of the concentration of NO. When concentration of NO is kept constant and concentration of O2 is doubled the rate also gets doubled indicating that rate depends on concentration of O2 to the first power. Hence, the rate equation for this reaction will be Rate = k [NO] 2[O2] 67 Chemical Kinetics Reprint 2025-26 The differential form of this rate expression is given as d [ R ] 2 − = k [ NO ] [ O 2 ] d t Now, we observe that for this reaction in the rate equation derived from the experimental data, the exponents of the concentration terms are the same as their stoichiometric coefficients in the balanced chemical equation. Some other examples are given below: Reaction Experimental rate expression 1. CHCl3 + Cl2 ® CCl4 + HCl Rate = k [CHCl3 ] [Cl2]1/2 2. CH3COOC2H5 + H2O ® CH3COOH + C2H5OH Rate = k [CH3COOC2H5]1 [H2O]0 In these reactions, the exponents of the concentration terms are not the same as their stoichiometric coefficients. Thus, we can say that: Rate law for any reaction cannot be predicted by merely looking at the balanced chemical equation, i.e., theoretically but must be determined experimentally. 3.2.3 Order of a In the rate equation (3.4) Reaction Rate = k [A]x [B]y x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Order of a reaction can be 0, 1, 2, 3 and even a fraction. A zero order reaction means that the rate of reaction is independent of the concentration of reactants. ExampleExampleExampleExampleExample 3.33.33.33.33.3 Calculate the overall order of a reaction which has the rate expression (a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 SolutionSolutionSolutionSolutionSolution (a) Rate = k [A]x [B]y order = x + y So order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order. A balanced chemical equation never gives us a true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary reactions. When a sequence of elementary reactions (called mechanism) gives us the products, the reactions are called complex reactions. Chemistry 68 Reprint 2025-26 These may be consecutive reactions (e.g., oxidation of ethane to CO2 and H2O passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed), reverse reactions and side reactions (e.g., nitration of phenol yields o-nitrophenol and p-nitrophenol). Units of rate constant For a general reaction aA + bB ® cC + dD Rate = k [A]x [B]y Where x + y = n = order of the reaction Rate k = x [A] [B]y concentration 1 = × n ( where [A] = [B]) time ( concentration ) Taking SI units of concentration, mol L –1 and time, s, the units of k for different reaction order are listed in Table 3.3 Table 3.3: Units of rate constant Reaction Order Units of rate constant mol L−1 1 −1 − 1 × 0 = mol L s −1 Zero order reaction 0 s ( mol L ) −1 mol L 1 −1 × = s − 1 1 First order reaction 1 s ( mol L ) − 1 mol L 1 − 1 −1 × = mol L s −1 2 Second order reaction 2 s ( mol L ) Identify the reaction order from each of the following rate constants. ExampleExampleExampleExampleExample 3.43.43.43.43.4 (i) k = 2.3 × 10–5 L mol–1 s–1 (ii) k = 3 × 10–4 s–1 (i) The unit of second order rate constant is L mol–1 s–1, therefore SolutionSolutionSolutionSolutionSolution k = 2.3 × 10–5 L mol–1 s–1 represents a second order reaction. (ii) The unit of a first order rate constant is s–1 therefore k = 3 × 10–4 s–1 represents a first order reaction. 3.2.4 Molecularity Another property of a reaction called molecularity helps in of a understanding its mechanism. The number of reacting species Reaction (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. The reaction can be unimolecular when one reacting species is involved, for example, decomposition of ammonium nitrite. 69 Chemical Kinetics Reprint 2025-26 NH4NO2 ® N2 + 2H2O Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide. 2HI ® H2 + I2 Trimolecular or termolecular reactions involve simultaneous collision between three reacting species, for example, 2NO + O2 ® 2NO2 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 + 3H2SO4 ® KCl + 3Fe2(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. Consider the decomposition of hydrogen peroxide which is catalysed by iodide ion in an alkaline medium. -I 2H2O2  2H2O + O2 Alkaline medium The rate equation for this reaction is found to be d  H 2 O 2   Rate  k  H2 O 2 I dt This reaction is first order with respect to both H2O2 and I–. Evidences suggest that this reaction takes place in two steps (1) H2O2 + I– ® H2O + IO– (2) H2O2 + IO– ® H2O + I– + O2 Both the steps are bimolecular elementary reactions. Species IO- is called as an intermediate since it is formed during the course of the reaction but not in the overall balanced equation. The first step, being slow, is the rate determining step. Thus, the rate of formation of intermediate will determine the rate of this reaction. Thus, from the discussion, till now, we conclude the following: (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. Chemistry 70 Reprint 2025-26 (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. IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.3 For a reaction, A + B ® Product; the rate law is given by, r = k [ A]1/2 [B]2. What is the order of the reaction? 3.4 The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y ? 3.33.33.33.33.3 IntegratedIntegratedIntegratedIntegratedIntegrated We have already noted that the concentration dependence of rate is RateRateRateRateRate called differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by determination EquationsEquationsEquationsEquationsEquations of slope of the tangent at point ‘t’ in concentration vs time plot (Fig. 3.1). This makes it difficult to determine the rate law and hence the order of the reaction. In order to avoid this difficulty, we can integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions. 3.3.1 Zero Order Zero order reaction means that the rate of the reaction is proportional Reactions to zero power of the concentration of reactants. Consider the reaction, R ® P d  R  Rate =   k  R 0 d t As any quantity raised to power zero is unity d  R  Rate =   k × 1 d t d[R] = – k dt Integrating both sides [R] = – k t + I (3.5) where, I is the constant of integration. At t = 0, the concentration of the reactant R = [R]0, where [R]0 is initial concentration of the reactant. Substituting in equation (3.5) [R]0 = –k × 0 + I [R]0 = I Substituting the value of I in the equation (3.5) [R] = -kt + [R]0 (3.6) 71 Chemical Kinetics Reprint 2025-26 Comparing (3.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight [R0 ] line (Fig. 3.3) with slope = –k and intercept equal to [R]0. Further simplifying equation (3.6), we get the rateR k = -slope constant, k as of [ R ]0 − [ R ] k = (3.7) t Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme Concentration catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order 0 Time reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at Fig. 3.3: Variation in the concentration high pressure. vs time plot for a zero order 1130K reaction 2NH 3 ( g ) →Pt catalyst N 2 ( g ) +3H 2 ( g ) Rate = k [NH3]0 = k In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration. The thermal decomposition of HI on gold surface is another example of zero order reaction. 3.3.2 First Order In this class of reactions, the rate of the reaction is proportional to the Reactions first power of the concentration of the reactant R. For example, R ® P d [ R ] Rate = − = k [ R ] d t d [ R ] or = – kdt [ R ] Integrating this equation, we get ln [R] = – kt + I (3.8) Again, I is the constant of integration and its value can be determined easily. When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (3.8) can be written as ln [R]0 = –k × 0 + I ln [R]0 = I Substituting the value of I in equation (3.8) ln[R] = –kt + ln[R]0 (3.9) Chemistry 72 Reprint 2025-26 Rearranging this equation [ R ] ln = −kt [ R ]0 1 R 0 or k  ln (3.10) t R At time t1 from equation (3.8) *ln[R]1 = – kt1 + *ln[R]0 (3.11) At time t2 ln[R]2 = – kt2 + ln[R]0 (3.12) where, [R]1 and [R]2 are the concentrations of the reactants at time t1 and t2 respectively. Subtracting (3.12) from (3.11) ln[R]1– ln[R]2 = – kt1 – (–kt2) [ R ]1 ln = k (t 2 − t 1 ) [ R ]2 1 [ R ]1 k = ln (t 2 − t 1 ) [ R ]2 (3.13) Equation (3.9) can also be written as [ R ] ln = −kt [ R ]0 Taking antilog of both sides [R] = [R]0 e–kt (3.14) Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]0 The first order rate equation (3.10) can also be written in the form 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 kt * log = [ R ] 2.303 If we plot a graph between log [R]0/[R] vs t, (Fig. 3.5), the slope = k/2.303 Hydrogenation of ethene is an example of first order reaction. C2H4(g) + H2 (g) ® C2H6(g) Rate = k [C2H4] All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. * Refer to Appendix-IV for ln and log (logarithms). 73 Chemical Kinetics Reprint 2025-26 /[R]) 0] Slope = k /2.303 ([R log 0 Time Fig. 3.4: A plot between ln[R] and t Fig. 3.5: Plot of log [R]0/[R] vs time for a for a first order reaction first order reaction 226 88 Ra  24 He  22286 Rn Rate = k [Ra] Decomposition of N2O5 and N2O are some more examples of first order reactions. ExampleExampleExampleExampleExample 3.53.53.53.53.5 The initial concentration of N2O5 in the following first order reaction N2O5(g) ® 2 NO2(g) + 1/2O2 (g) was 1.24 × 10–2 mol L–1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 × 10–2 mol L–1. Calculate the rate constant of the reaction at 318 K. SolutionSolutionSolutionSolutionSolution For a first order reaction  R 1 k t 2  t 1  log =  R 2 2.303 2.303  R 1 log k =  t 2  t 1   R 2 2.303 1.24  10  2 mol L1 log  2  1 =  60 min  0 min  0.20  10 mol L 2.303  1 = log 6.2 min 60 k = 0.0304 min-1 Let us consider a typical first order gas phase reaction A(g) ® B(g) + C(g) Let pi be the initial pressure of A and pt the total pressure at time ‘t’. Integrated rate equation for such a reaction can be derived as Total pressure pt = pA + pB + pC (pressure units) Chemistry 74 Reprint 2025-26 pA, pB and pC are the partial pressures of A, B and C, respectively. If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each. A(g) ® B(g) + C(g) At t = 0 pi atm 0 atm 0 atm At time t (pi–x) atm x atm x atm where, pi is the initial pressure at time t = 0. pt = (pi – x) + x + x = pi + x x = (pt - pi) where, pA = pi – x = pi – (pt – pi) = 2pi – pt  2.303   p i  k =   log  (3.16)  t  p A  2.303 p i log = t  2 p i  p t  The following data were obtained during the first order thermal ExampleExampleExampleExampleExample 3.63.63.63.63.6 decomposition of N2O5 (g) at constant volume: 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) S.No. Time/s Total Pressure/(atm) 1. 0 0.5 2. 100 0.512 Calculate the rate constant. Let the pressure of N2O5(g) decrease by 2x atm. As two moles of SolutionSolutionSolutionSolutionSolution N2O5 decompose to give two moles of N2O4(g) and one mole of O2 (g), the pressure of N2O4 (g) increases by 2x atm and that of O2 (g) increases by x atm. 2N 2 O5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g ) Start t = 0 0.5 atm 0 atm 0 atm At time t (0.5 – 2x) atm 2x atm x atm pt = p N 2 O 5  p N 2 O 4  p O 2 = (0.5 – 2x) + 2x + x = 0.5 + x x = tp − 0.5 p N 2 O5 = 0.5 – 2x = 0.5 – 2 (pt – 0.5) = 1.5 – 2pt At t = 100 s; pt = 0.512 atm 75 Chemical Kinetics Reprint 2025-26 p N 2 O 5 = 1.5 – 2 × 0.512 = 0.476 atm Using equation (3.16) 2.303 p i 2.303 0.5 atm k  log  log t p A 100s 0.476 atm 2.303  4 1   0.0216  4.98  10 s 100s 3.3.3 Half-Life of The half-life of a reaction is the time in which the concentration of a a Reaction reactant is reduced to one half of its initial concentration. It is represented as t1/2. For a zero order reaction, rate constant is given by equation 3.7. [ R ]0 − [ R ] k = t 1 [ R ]0 At t = t 1/2 , [ R ] = 2 The rate constant at t1/2 becomes [ R ]0 − 1/2 [ R ]0 k = t 1/2 [ R ]0 t 1/2 = 2k It is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant. For the first order reaction, 2.303 [ R ]0 k = log (3.15) t [ R ] [ R ]0 at t1/2 [ R ] = (3.16) 2 So, the above equation becomes 2.303 [ R ]0 k = log t 1/2 [ R ] /2 2.303 or t1/2  log 2 k 2.303 t 1/2 = × 0.301 k 0.693 t 1/2 = (3.17) k Chemistry 76 Reprint 2025-26 It can be seen that for a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species. The half-life of a first order equation is readily calculated from the rate constant and vice versa. For zero order reaction t1/2 µ [R]0. For first order reaction t1/2 is independent of [R]0. A first order reaction is found to have a rate constant, k = 5.5 × 10-14 s-1. ExampleExampleExampleExampleExample 3.73.73.73.73.7 Find the half-life of the reaction. Half-life for a first order reaction is SolutionSolutionSolutionSolutionSolution 0.693 t 1/2 = k 0.693 t 1/2 = –14 –1 = 1.26 × 1013s 5.5×10 s Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2) of the reaction. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0 ExampleExampleExampleExampleExample 3.83.83.83.83.8 2.303  R 0 log k = SolutionSolutionSolutionSolutionSolution t  R  2.303  R 0 2.303 3 log = = log10 t  R 0  0.999  R 0 t t = 6.909/k For half-life of the reaction t1/2 = 0.693/k t 6.909 k =   10 t1/2 k 0.693 Table 3.4 summarises the mathematical features of integrated laws of zero and first order reactions. Table 3.4: Integrated Rate Laws for the Reactions of Zero and First Order Order Reaction Differential Integrated Straight Half- Units of k type rate law rate law line plot life 0 R® P d[R]/dt = -k kt = [R]0-[R] [R] vs t [R]0/2k conc time-1 or mol L–1s–1 1 R® P d[R]/dt = -k[R] [R] = [R]0e-kt ln[R] vs t ln 2/k time-1 or s–1 or kt = ln{[R]0/[R]} 77 Chemical Kinetics Reprint 2025-26 The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. For example, during the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, amounts of the reactants and products at the beginning (t = 0) and completion (t) of the reaction are given as under. H CH3COOH + C2H5OH CH3COOC2H5 + H2O  t = 0 0.01 mol 10 mol 0 mol 0 mol t 0 mol 9.99 mol 0.01 mol 0.01 mol The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction. C12H22O11 + H2O →H+ C6H12O6 + C6H12O6 Cane sugar Glucose Fructose Rate = k [C12H22O11] IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.5 A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g of this reactant take to reduce to 3 g? 3.6 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. 3.43.43.43.43.4 TemperatureTemperatureTemperatureTemperatureTemperature Most of the chemical reactions are accelerated by increase in temperature. For example, in decomposition of N2O5, the time taken for half of the DependenceDependenceDependenceDependenceDependence ofofofofof original amount of material to decompose is 12 min at 50oC, 5 h at thethethethethe RateRateRateRateRate ofofofofof aaaaa 25oC and 10 days at 0oC. You also know that in a mixture of potassium ReactionReactionReactionReactionReaction permanganate (KMnO4) and oxalic acid (H2C2O4), potassium permanganate gets decolourised faster at a higher temperature than that at a lower temperature. It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation (3.18). It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation. Chemistry 78 Reprint 2025-26 k = A e -Ea /RT (3.18) where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol –1). It can be understood clearly using the following simple reaction H 2 g  I 2 g  2HI g According to Arrhenius, this reaction can take place only when a molecule of hydrogen and a molecule of iodine Intermediate collide to form an unstable intermediate (Fig. 3.6). It exists for a very short time and then breaks up to form two Fig. 3.6: Formation of HI through molecules of hydrogen iodide. the intermediate The energy required to form this intermediate, called activated complex (C), is known as activation energy (Ea). Fig. 3.7 is obtained by plotting potential energy vs reaction coordinate. Reaction coordinate represents the profile of energy change when reactants change into products. Some energy is released when the complex decomposes to form products. So, the final enthalpy of the reaction depends upon the nature of reactants and products. All the molecules in the reacting species do not have the same kinetic Fig. 3.7: Diagram showing plot of potential energy. Since it is difficult to predict the energy vs reaction coordinate behaviour of any one molecule with precision, Ludwig Boltzmann and James Clark Maxwell used statistics to predict the behaviour of large number of molecules. According to them, the distribution of kinetic energy may be described by plotting the fraction of molecules (NE/NT) with a given kinetic energy (E) vs kinetic energy (Fig. 3.8). Here, NE is the number of molecules with energy E and NT is total number of molecules. The peak of the curve corresponds to the most probable kinetic energy, i.e., kinetic energy of maximum fraction of molecules. There are decreasing number Fig. 3.8: Distribution curve showing energies of molecules with energies higher or among gaseous molecules lower than this value. When the 79 Chemical Kinetics Reprint 2025-26 temperature is raised, the maximum of the curve moves to the higher energy value (Fig. 3.9) and the curve broadens out, i.e., spreads to the right such that there is a greater proportion of molecules with much higher energies. The area under the curve must be constant since total probability must be one at all times. We can mark the position of Ea on Fig. 3.9: Distribution curve showing temperature Maxwell Boltzmann distribution curve dependence of rate of a reaction (Fig. 3.9). Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than Ea. It is clear from the diagram that in the curve at (t + 10), the area showing the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of a reaction. In the Arrhenius equation (3.18) the factor e -Ea /RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) E a ln k = – + ln A (3.19) RT The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. E a In Fig. 3.10, slope = – and intercept = ln R A. So we can calculate Ea and A using these values. At temperature T1, equation (3.19) is E a ln k1 = – RT1 + ln A (3.20) At temperature T2, equation (3.19) is E a ln k2 = – RT2 + ln A (3.21) (since A is constant for a given reaction) k1 and k2 are the values of rate constants at temperatures T1 and T2 respectively. Fig. 3.10: A plot between ln k and 1/T Chemistry 80 Reprint 2025-26 Subtracting equation (3.20) from (3.21), we obtain E a E a ln k2 – ln k1 = RT1 – RT2 k 2 E a  1 1  ln = − k1 R  T1 T2  k 2 E a  1 1  log = − (3.22) k1 2.303 R  T1 T2  k 2 E a  T2 − T1  log = k1 2. 303R  T1T2  ExampleExampleExampleExampleExample 3.93.93.93.93.9 The rate constants of a reaction at 500K and 700K are 0.02s–1 and 0.07s–1 respectively. Calculate the values of Ea and A. k 2 E a  T2  T1 SolutionSolutionSolutionSolutionSolution log =   k1 2.303 R  T1T2  0.07  E a   700  500  log =      700  500 0.02  2.303  8.314 JK  1 mol  1  0.544 = Ea × 5.714 × 10-4/19.15 Ea = 0.544 × 19.15/5.714 × 10–4 = 18230.8 J Since k = Ae-Ea/RT × 500 0.02 = Ae-18230.8/8.314 A = 0.02/0.012 = 1.61 ExampleExampleExampleExampleExample 3.103.103.103.103.10 The first order rate constant for the decomposition of ethyl iodide by the reaction C2H5I(g) ® C2H4 (g) + HI(g) at 600K is 1.60 × 10–5 s–1. Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700K. SolutionSolutionSolutionSolutionSolution We know that E a  1 1     log k2 – log k1 = 2.303R  T1 T2  81 Chemical Kinetics Reprint 2025-26 E a  1 1  log k2 = log k1     2.303R  T1 T2  209000 J mol L 1  1 1  5 = log 1.60  10     1  1    2.303  8.314 J mol L K  600 K 700K  log k2 = – 4.796 + 2.599 = – 2.197 k2 = 6.36 × 10–3 s–1 3.4.1 Effect of A catalyst is a substance which increases the rate of a reaction without Catalyst 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. 3.11. It is clear from Arrhenius equation (3.18) 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 Fig. 3.11: Effect of catalyst on activation energy not alter Gibbs energy, DG of a reaction. It catalyses the spontaneous reactions but does not catalyse non-spontaneous 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. 3.53.53.53.53.5 CollisionCollisionCollisionCollisionCollision Though Arrhenius equation is applicable under a wide range of TheoryTheoryTheoryTheoryTheory ofofofofof circumstances, collision theory, which was developed by Max Trautz ChemicalChemicalChemicalChemicalChemical and William Lewis in 1916 -18, provides a greater insight into the energetic and mechanistic aspects of reactions. It is based on kinetic ReactionsReactionsReactionsReactionsReactions theory of gases. According to this theory, the reactant molecules are Chemistry 82 Reprint 2025-26 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 a / RT (3.23) Rate = Z AB e − E 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 (3.23) with Arrhenius equation, we can say that A is related to collision frequency. Equation (3.23) 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. For example, formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown in Fig. 3.12. The proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed. Fig. 3.12: Diagram showing molecules having proper and To account for effective collisions, improper orientation another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e., − E a / RT Rate = PZ AB e Thus, in collision theory activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. Collision theory also has certain drawbacks as it considers atoms/ molecules to be hard spheres and ignores their structural aspect. You will study details about this theory and more on other theories in your higher classes. * Threshold energy = Activation Energy + energy possessed by reacting species. 83 Chemical Kinetics Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 3.7 What will be the effect of temperature on rate constant ? 3.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. 3.9 The activation energy for the reaction 2 HI(g) ® H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? SummarySummarySummarySummarySummary Chemical kinetics is the study of chemical reactions with respect to reaction rates, effect of various variables, rearrangement of atoms and formation of intermediates. The rate of a reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. The order of a reaction is the sum of all such powers of concentration of terms for different reactants. Rate constant is the proportionality factor in the rate law. Rate constant and order of a reaction can be determined from rate law or its integrated rate equation. Molecularity is defined only for an elementary reaction. Its values are limited from 1 to 3 whereas order can be 0, 1, 2, 3 or even a fraction. Molecularity and order of an elementary reaction are same. Temperature dependence of rate constants is described by Arrhenius equation (k = Ae–Ea/RT). Ea corresponds to the activation energy and is given by the energy difference between activated complex and the reactant molecules, and A (Arrhenius factor or pre-exponential factor) corresponds to the collision frequency. The equation clearly shows that increase of temperature or lowering of Ea will lead to an increase in the rate of reaction and presence of a catalyst lowers the activation energy by providing an alternate path for the reaction. According to collision theory, another factor P called steric factor which refers to the orientation of molecules which collide, is important and contributes to effective collisions, thus, modifying the Arrhenius equation to k  P Z AB e  E a / RT . Chemistry 84 Reprint 2025-26 ExercisesExercisesExercisesExercisesExercises 3.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) ® N2O (g) Rate = k[NO]2 (ii) H2O2 (aq) + 3I– (aq) + 2H+ ® 2H2O (l) + 3I Rate = k[H2O2][I-] (iii) CH3CHO (g) ® CH4 (g) + CO(g) Rate = k [CH3CHO]3/2 (iv) C2H5Cl (g) ® C2H4 (g) + HCl (g) Rate = k [C2H5Cl] 3.2 For the reaction: 2A + B ® A2B the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. 3.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1? 3.4 The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., 3/2 Rate = k ( p CH 3 OCH 3 ) If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? 3.5 Mention the factors that affect the rate of a chemical reaction. 3.6 A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ? 3.7 What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively? 3.8 In a pseudo first order reaction in water, the following results were obtained: t/s 0 30 60 90 [A]/ mol L–1 0.55 0.31 0.17 0.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.