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Unit, , Objectives, , 4, , Chemical K, ine, tics, Kine, inetics, , After studying this Unit, you will be, able to, •, •, , •, •, , •, •, , •, , •, •, , define, the, average, and, instantaneous rate of a reaction;, express the rate of a reaction in, terms of change in concentration, of either of the reactants or, products with time;, distinguish between elementary, and complex reactions;, differentiate, between, the, molecularity and order of a, reaction;, define rate constant;, discuss the dependence of rate of, reactions, on, concentration,, temperature and catalyst;, derive integrated rate equations, for the zero and first order, reactions;, determine the rate constants for, zeroth and first order reactions;, describe collision theory., , Chemical Kinetics helps us to understand how chemical reactions, occur., , Chemistry, by its very nature, is concerned with change., Substances with well defined properties are converted, by chemical reactions into other substances with, different properties. For any chemical reaction, chemists, try to find out, (a) the feasibility of a chemical reaction which can be, predicted by thermodynamics ( as you know that a, reaction with ΔG < 0, at constant temperature and, pressure is feasible);, (b) extent to which a reaction will proceed can be, determined from chemical equilibrium;, (c) speed of a reaction i.e. time taken by a reaction to, reach equilibrium., Along with feasibility and extent, it is equally, important to know the rate and the factors controlling, the rate of a chemical reaction for its complete, 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

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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., , 4.1 Rate of a, Chemical, Reaction, , Some reactions such as ionic reactions occur very fast, for example,, precipitation of silver chloride occurs instantaneously by mixing of, aqueous solutions of silver nitrate and sodium chloride. On the other, 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,, Δt, , = t2 – t1, , Δ[R] = [R]2 – [R]1, Δ [P] = [P]2 – [P]1, The square brackets in the above expressions are used to express, molar concentration., Rate of disappearance of R, =, , Chemistry, , 94, , Δ [R ], Decrease in concentration of R, =−, Time taken, Δt, , (4.1)

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Rate of appearance of P, Increase in concentration of P, Δ [P], =+, (4.2), Time taken, Δt, Since, Δ[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 (4.1) and (4.2) 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. 4.1)., =, , Concentration of reactants, , [R]0, , c1, , rav, , D[R], , c2, , =, , D [R], Dt, , =, , (c2–c1), (t2–t1), , d[R], Dt, , d [R], , rinst = d t, , =, , slope, , dt, , t1, , t2, , t, , Time, (a), , Fig. 4.1: Instantaneous and average rate of a reaction, Units of rate of a reaction, , From equations (4.1) and (4.2), 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, below, calculate the average rate of the reaction:, , Example 4.1, , C4H9Cl + H2O → C4H9OH + HCl, during different intervals of time., t/s, 0, 50, 100, 150, 200, 300, 400, 700, 800, –1, [C4H9Cl]/mol L 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, of time and thus determine the average rate by dividing Δ[R] by Δt, (Table 4.1)., , 95, , Solution, , Chemical Kinetics

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Table 4.1: Average rates of hydrolysis of butyl chloride, [C4H9CI]t /, , mol L–1, , mol L–1, , 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, , 1, , 2, , t 1 /s, , t 2 /s, , rav × 104/mol L–1s–1, , [C4H9CI]t /, , =–, , {[C H Cl], 4, , 9, , t2, , }, , – [ C 4 H 9 Cl ]t / ( t 2 − t1 ) × 10 4, 1, , It can be seen (Table 4.1) 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, Δt approaches zero). Hence, mathematically for an infinitesimally small, dt instantaneous rate is given by, rav =, , −Δ [ R ] Δ [ P ], =, Δt, Δt, , As Δt → 0, , Fig 4.2, Instantaneous rate of, hydrolysis of butyl, chloride(C4H9Cl), , Chemistry, , 96, , or, , (4.3), rinst =, , −d [ R ] d [ P ], =, dt, dt

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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. 4.1). So in problem 4.1, 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. 4.2)., The slope of this tangent gives the instantaneous rate., ⎛ 0.0165 – 0.037 ⎞, So, rinst at 600 s = ⎜ ( 800 – 400) 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, Rate of reaction = –, , Δ [ Hg ], Δ [ Cl 2 ] Δ [ HgCl 2 ], =–, =, Δt, Δt, Δt, , 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 Δ[HI] is divided by 2. The rate of this reaction, is given by, Rate of reaction = −, , 1 Δ [ HI] Δ [ H2 ] Δ [ I2 ], =, =, 2 Δt, Δt, Δt, , Similarly, for the reaction, 5 Br- (aq) + BrO3– (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l), , Rate = −, , Δ ⎡BrO3− ⎤⎦, 1 Δ [ Br − ], 1 Δ [ H+ ] 1 Δ [ Br2 ] 1 Δ [ H2O], =− ⎣, =−, =, =, 5 Δt, 6 Δt, 3 Δt, 3, Δt, Δt, , 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., , 97, , Chemical Kinetics

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Example 4.2 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?, , Solution, , Average Rate =, , 1 ⎧ Δ [ N 2 O5 ] ⎫, 1 ⎡ ( 2.08 − 2.33) mol L−1 ⎤, ⎨−, ⎬=− ⎢, ⎥, 2⎩, 2⎣, 184 min, Δt, ⎭, ⎦, , = 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, Rate =, , 1 ⎧ Δ [ NO2 ] ⎫, ⎨, ⎬, 4 ⎩ Δt ⎭, , Δ [ NO2 ], = 6.79 × 10–4 × 4 mol L–1 min–1 = 2.72 × 10–3 mol L–1min–1, Δt, , Intext Questions, 4.1 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., 4.2 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?, , 4.2 Factors Influencing, Rate of a Reaction, , Rate of reaction depends upon the experimental conditions such, as concentration of reactants (pressure in case of gases),, temperature and catalyst., , 4.2.1 Dependence, of Rate on, Concentration, , The rate of a chemical reaction at a given temperature may depend on, the concentration of one or more reactants and products. The, 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., , 4.2.2 Rate, Expression, and Rate, Constant, , The results in Table 4.1 clearly show that rate of a reaction decreases with, the passage of time as the concentration of reactants decrease. Conversely,, rates generally increase when reactant concentrations increase. So, rate of, a reaction depends upon the concentration of reactants., , Chemistry, , 98

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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, x, y, Rate ∝ [A] [B], (4.4), 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, x, y, Rate = k [A] [B], (4.4a), d [R ], −, = k [ A ]x [ B] y, (4.4b), dt, This form of equation (4.4 b) is known as differential rate equation,, where k is a proportionality constant called rate constant. The equation, like (4.4), 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 4.2)., Table 4.2: 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, 2, Rate = k[NO] [O2], 99, , Chemical Kinetics

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The differential form of this rate expression is given as, d [R ], −, = k [ NO ]2 [ O2 ], dt, 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, 1, 0, 2. CH3COOC2H5 + H2O → CH3COOH + C2H5OH Rate = k [CH3COOC2H5] [H2O], 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., 4.2.3 Order of a, Reaction, , In the rate equation (4.4), 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 (4.4) 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., , Example 4.3 Calculate the overall order of a reaction which, Solution, , has the rate expression, (a) Rate = k [A]1/2 [B]3/2, (b) Rate = k [A]3/2 [B]–1, , (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 100

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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 =, [A]x [B]y, 1, = concentration ×, n, time, ( concerntration ), –1, , Taking SI units of concentration, mol L and time, s, the units of, k for different reaction order are listed in Table 4.3, Table 4.3: Units of rate constant, Reaction, Zero order reaction, , First order reaction, , Second order reaction, , Order, , Units of rate constant, , 0, , mol L−1, 1, ×, = mol L−1s −1, 0, s, ( mol L−1 ), , 1, , mol L−1, 1, ×, = s −1, −1 1, s, ( mol L ), , 2, , mol L−1, 1, ×, = mol −1L s −1, −1 2, s, ( mol L ), , Identify the reaction order from each of the following rate constants., (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, 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., , 4.2.4 Molecularity, of a, Reaction, , Example 4.4, , Solution, , Another property of a reaction called molecularity helps in, The number of reacting species, understanding its mechanism., (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., 101 Chemical Kinetics

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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, the molecularity greater, than three is not observed., It is, therefore, evident that complex reactions involving more than, three molecules in the stoichiometric equation must take place in more, than one step., 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-, , → 2H2O + O2, 2H2O2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯, Alkaline medium, , The rate equation for this reaction is found to be, −d [ H 2 O 2 ], , = k [ H2O2 ] [I− ], dt, This reaction is first order with respect to both H2O2 and I–. Evidences, suggest that this reaction takes place in two steps, Rate =, , (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 102

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(iii) For complex reaction, order is given by the slowest step and, generally, molecularity of the slowest step is same as the order of, the overall reaction., , Intext Questions, 4.3 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?, 4.4 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 ?, , 4 . 3 Integrated, Rate, Equations, , We have already noted that the concentration dependence of rate is, called differential rate equation. It is not always convenient to, determine the instantaneous rate, as it is measured by determination, of slope of the tangent at point ‘t’ in concentration vs time plot, (Fig. 4.1). 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., , 4.3.1 Zero Order, Reactions, , Zero order reaction means that the rate of the reaction is proportional, to zero power of the concentration of reactants. Consider the reaction,, R → P, d [R ], = k [ R ]0, Rate = −, dt, As any quantity raised to power zero is unity, d [R ], =k ×1, Rate = −, dt, d[R] = – k dt, Integrating both sides, [R] = – k t + I, (4.5), 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 (4.5), [R]0 = –k × 0 + I, [R]0 = I, Substituting the value of I in the equation (4.5), [R] = -kt + [R]0, , (4.6), , 103 Chemical Kinetics

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Comparing (4.6) with equation of a straight line,, y = mx + c, if we plot [R] against t, we get a straight, line (Fig. 4.3) with slope = –k and intercept equal, to [R]0., Further simplifying equation (4.6), we get the rate, constant, k as, , Concentration of R, , [R0], , 0, , k = -slope, , k=, , Time, , Fig. 4.3: Variation in the concentration vs, time plot for a zero order reaction, , [ R ]0 − [ R ], t, , (4.7), , Zero order reactions are relatively uncommon but, they occur under special conditions. Some enzyme, catalysed reactions and reactions which occur on, metal surfaces are a few examples of zero order, reactions. The decomposition of gaseous ammonia, on a hot platinum surface is a zero order reaction at, high pressure., 1130K, 2NH3 ( g ) ⎯⎯⎯⎯⎯, → N 2 ( g ) +3H2 ( g ), Pt catalyst, , 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., 4.3.2 First Order, Reactions, , In this class of reactions, the rate of the reaction is proportional to the, first power of the concentration of the reactant R. For example,, R → P, Rate = −, , or, , d [R ], = k [R ], dt, , d [R ], = –kdt, [R ], , Integrating this equation, we get, ln [R] = – kt + I, , (4.8), , 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 (4.8) can be written as, ln [R]0 = –k × 0 + I, ln [R]0 = I, Substituting the value of I in equation (4.8), ln[R] = -kt + ln[R]0, Chemistry 104, , (4.9)

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Rearranging this equation, ln, , [R ], = −kt, [ R ]0, , or k =, , [R ]0, 1, ln, t, [R ], , (4.10), , At time t1 from equation (4.8), *ln[R]1 = – kt1 + *ln[R]0, , (4.11), , At time t2, ln[R]2 = – kt2 + ln[R]0, , (4.12), , where, [R]1 and [R]2 are the concentrations of the reactants at time, t1 and t2 respectively., Subtracting (4.12) from (4.11), ln[R]1– ln[R]2 = – kt1 – (–kt2), ln, , [ R ]1, = k (t 2 − t1 ), [ R ]2, , k=, , 1, [ R ]1, ln, (t 2 − t1 ) [ R ]2, , (4.13), , Equation (4.9) can also be written as, ln, , [R ], = −kt, [ R ]0, , Taking antilog of both sides, [R] = [R]0 e-kt, , (4.14), , Comparing equation (4.9) with y = mx + c, if we plot ln [R] against, t (Fig. 4.4) we get a straight line with slope = –k and intercept equal to, ln [R]0, The first order rate equation (4.10) can also be written in the form, k=, , [ R ]0, 2.303, log, t, [R ], , (4.15), , [ ], kt, * log R 0 =, [ R ] 2.303, If we plot a graph between log [R]0/[R] vs t, (Fig. 4.5),, 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)., , 105 Chemical Kinetics

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log ([R]0/[R]), , Slope = k/2.303, , 0, , Fig. 4.4: A plot between ln[R] and t, for a first order reaction, 226, 88, , Ra → 42 He +, , Time, , Fig. 4.5: Plot of log [R]0/[R] vs time for a first, order reaction, 222, 86, , Rn, , Rate = k [Ra], Decomposition of N2O5 and N2O are some more examples of first, order reactions., , Example 4.5 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., , Solution For a first order reaction, log, , [R ]1, [R ]2, , =, , k (t 2 − t1 ), , 2.303, [R ]1, 2.303, k = (t − t ) log [ ], R 2, 2, 1, 1.24 × 10 −2 mol L−1, 2.303, log, =, ( 60 min− 0 min ) 0.20 × 10−2 mol L−1, =, , 2.303, log 6.2 min −1, 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 106

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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, x atm, x atm, At time t, (pi–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 ⎞ ⎛ log p i ⎞, ⎟⎜, k = ⎜, p A ⎟⎠, ⎝ t ⎠⎝, =, , (4.16), , 2.303, pi, log, p, 2, t, ( i − pt ), , The following data were obtained during the first order thermal, decomposition of N2O5 (g) at constant volume:, , Example 4.6, , 2N2 O5 ( g ) → 2N 2 O4 ( g ) + O2 ( g ), S.No., Time/s, 1., 0, 2., 100, Calculate the rate constant., , Total Pressure/(atm), 0.5, 0.512, , Let the pressure of N2O5(g) decrease by 2x atm. As two moles of, 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., , 2N2 O5 ( g ), Start t = 0, At time t, , →, , 0.5 atm, (0.5 – 2x) atm, , 2N 2 O4 ( g ), 0 atm, 2x atm, , +, , Solution, , O2 ( g ), 0 atm, x atm, , pt = p N2O5 + p N2O4 + pO2, = (0.5 – 2x) + 2x + x = 0.5 + x, x = pt − 0.5, , pN2O5, At, , = 0.5 – 2x, = 0.5 – 2 (pt – 0.5) = 1.5 – 2pt, t = 100 s; pt = 0.512 atm, 107 Chemical Kinetics

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p N 2 O5, , = 1.5 – 2 × 0.512 = 0.476 atm, , Using equation (4.16), 2.303, 2.303, 0.5 atm, p, log i =, log, 100 s, 0.476 atm, t, pA, , k=, =, , 4.3.3 Half-Life of a, Reaction, , 2.303, × 0.0216 = 4.98 × 10−4 s −1, 100 s, , The half-life of a reaction is the time in which the concentration of a, 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 4.7., k=, , [ R ]0 − [ R ], t, , At t = t1/ 2 , [ R ] =, , 1 [ R ]0, 2, , The rate constant at t1/2 becomes, k =, , [ R ]0 − 1/ 2 [ R ]0, t1/ 2, , t1/ 2 =, , [ R ]0, , 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,, k=, , 2.303, [ R ]0, log, t, [R ], , at t1/2, , [R ] =, , [ R ]0, , 2, So, the above equation becomes, , k=, , or, , (4.16), , 2.303, [ R ]0, log, t1/ 2, [R ]/ 2, 2.303, log 2, k, 2.303, =, × 0.301, k, 0.693, =, k, , t1/ 2 =, t1/ 2, t1/ 2, , Chemistry 108, , (4.15), , (4.17)

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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 ∝ [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., Find the half-life of the reaction., , Example 4.7, Solution, , Half-life for a first order reaction is, 0.693, k, 0.693, =, = 1.26 × 1014s, 5.5 ×10 –14 s –1, , t1/ 2 =, t1/ 2, , 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, k =, , =, , [ R ]0, 2.303, log, [R ], t, , Example 4.8, Solution, , [ R ]0, 2.303, 2.303, log, log103, t, [ R ]0 − 0.999 [ R ]0 =, t, , t = 6.909/k, For half-life of the reaction, t1/2 = 0.693/k, , t, t1/ 2, , =, , 6.909, k, ×, = 10, k, 0.693, Table 4.4 summarises the mathematical features of integrated laws of, zero and first order reactions., , Table 4.4: Integrated Rate Laws for the Reactions of Zero and First Order, Order, , Reaction, type, , Differential, rate law, , Integrated, rate law, , Straight, line plot, , Halflife, , Units of k, , 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, or kt =, ln{[R]0/[R]}, , ln[R] vs t, , ln 2/k, , time-1 or s–1, , 109 Chemical Kinetics

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4.4 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 (t = 0) and completion (t) of the reaction are given, as under., +, , H, ⎯⎯⎯, →, , CH3COOC2H5 + H2O, t = 0, t, , 0.01 mol, 0 mol, , CH3COOH, , 10 mol, 9.9 mol, , 0 mol, 0.01 mol, , +, , C2H5OH, 0 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. The equation, thus, becomes, Rate = k [CH3COOC2H5], where k = k′ [H2O], , and 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, ⎯⎯⎯, →, , Cane sugar, , C6H12O6, Glucose, , +, , C6H12O6, Fructose, , Rate = k [C12H22O11], , Example 4.9 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, –1, , C/mol L, , 0, , 30, , 60, , 90, , 0.8500, , 0.8004, , 0.7538, , 0.7096, , Show that it follows a pseudo first order reaction, as the concentration, of water remains nearly constant (55 mol L–1), during the course of the, reaction. What is the value of k′ in this equation?, Rate = k′ [CH3COOCH3][H2O], , Solution For pseudo first order reaction, the reaction should be first order with, , respect to ester when [H2O] is constant. The rate constant k for pseudo, first order reaction is, k =, , Chemistry 110, , C, 2.303, log 0, t, C, , where k = k ' [ H2 O]

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From the above data we note, t/min, , C/ mol L–1, , k/mol L–1, , 0, , 0.8500, , –, , 30, , 0.8004, , 2.004 × 10–3, , 60, , 0.7538, , 2.002 × 10–3, , 90, , 0.7096, , 2.005 × 10–3, , It can be seen that k [H2O] is constant and equal to 2.004 × 10-3 min–1, and hence, it is pseudo first order reaction. We can now determine k from, k [H2O] = 2.004 × 10–3 min–1, k [55 mol L–1] = 2.004 × 10–3 min–1, k = 3.64 × 10–5 mol–1 L min–1, , Intext Questions, 4.5 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?, 4.6 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., , 4.5 Temperature, Dependence of, the Rate of a, Reaction, , Most of the chemical reactions are accelerated by increase in temperature., For example, in decomposition of N2O5, the time taken for half of the, original amount of material to decompose is 12 min at 50oC, 5 h at, 25oC and 10 days at 0oC. You also know that in a mixture of potassium, permanganate (KMnO 4) and oxalic acid (H 2C2O4), 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 (4.18). It was first, proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist,, Arrhenius provided its physical justification and interpretation., (4.18), k = A e -Ea /RT, 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)., , 111 Chemical Kinetics

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It can be understood clearly using the following simple reaction, H2 ( g ) + I2 ( g ) → 2HI ( g ), , According to Arrhenius, this reaction can take place only, when a molecule of hydrogen and a molecule of iodine collide, Intermediate, to form an unstable intermediate (Fig. 4.6). It exists for a very, Fig. 4.6: Formation of HI through short time and then breaks up to form two molecules of, hydrogen iodide., the intermediate, The energy required to form this, intermediate,, called, activated, complex (C), is known as activation, energy (Ea). Fig. 4.7 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 heat of the, reaction depends upon the nature of, reactants and products., Fig. 4.7: Diagram showing plot of potential, energy vs reaction coordinate., All the molecules in the reacting, species do not have the same kinetic, energy. Since it is difficult to predict, the 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, Fig. 4.8: Distribution curve showing energies, energy (Fig. 4.8). Here, NE is the, among gaseous molecules, 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 of molecules with, energies higher or lower than this, value. When the temperature is raised,, the maximum of the curve moves to, the higher energy value (Fig. 4.9) and, the curve broadens out, i.e., spreads, Fig. 4.9: Distribution curve showing temperature, to the right such that there is a greater, dependence of rate of a reaction, Chemistry 112

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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 Maxwell Boltzmann distribution, curve (Fig. 4.9)., 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 (4.18) 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 (4.18), ln k = –, , Ea, + ln A, RT, , (4.19), , The plot of ln k vs 1/T gives a straight line according to the equation, (4.19) as shown in Fig. 4.10., Thus, it has been found from Arrhenius equation (4.18) 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., Ea, and intercept = ln A. So we can, In Fig. 4.10, slope = –, R, calculate Ea and A using these values., At temperature T1, equation (4.19) is, Ea, ln k1 = – RT + ln A, 1, , (4.20), , At temperature T2, equation (4.19) is, Ea, (4.21), ln k2 = – RT + ln A, 2, (since A is constant for a given reaction), k1 and k2 are the values of rate constants at, temperatures T1 and T2 respectively., Subtracting equation (4.20) from (4.21), we obtain, Ea, Ea, ln k2 – ln k1 = RT – RT, 1, 2, , ln, , Fig. 4.10: A plot between ln k and 1/T, , k2 Ea, =, k1, R, , 1⎤, ⎡1, ⎢T − T ⎥, 2⎦, ⎣ 1, , log, , 1⎤, Ea, k2, ⎡1, − ⎥, =, ⎢, k1 2.303R ⎣ T1 T2 ⎦, , log, , k2, Ea, ⎡ T2 − T1 ⎤, =, k1 2.303R ⎢⎣ T1T2 ⎥⎦, , (4.22), , 113 Chemical Kinetics

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constants of a reaction at 500K and 700K are 0.02s–1 and, Example 4.10 The rate, –1, 0.07s respectively. Calculate the values of Ea and A., , Solution, , log, , log, , k2, k1, , =, , Ea, ⎡ T2 − T1 ⎤, 2.303R ⎢⎣ T1T2 ⎥⎦, , Ea, ⎛, ⎞ ⎡ 700 − 500 ⎤, 0.07, = ⎜ 2.303 × 8.314 JK −1mol −1 ⎟ ⎢, ⎥, 0.02, ⎝, ⎠ ⎣ 700 × 500 ⎦, , 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, 0.02 = Ae-18230.8/8.314 × 500, A = 0.02/0.012 = 1.61, , Example 4.11 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., , Solution We know that, log k2 – log k1 =, , Ea, 2.303R, , 1⎤, ⎡1, ⎢T − T ⎥, 2⎦, ⎣ 1, , log k2 = log k1 +, , Ea, 2.303R, , 1⎤, ⎡1, ⎢T − T ⎥, ⎣ 1, 2⎦, , = log (1.60 × 10−5 ) +, , 209000 J mol L−1, 2.303 × 8.314 J mol L−1K −1, , 1 ⎤, ⎡ 1, ⎢ 600 K − 700 K ⎥, ⎣, ⎦, , log k2 = – 4.796 + 2.599 = – 2.197, k2 = 6.36 × 10–3 s–1, , 4.5.1 Effect of, Catalyst, , A catalyst is a substance which alters 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., MnO2, 2 KCl + 3O2, 2KClO3 ⎯⎯⎯⎯→, 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., , Chemistry 114

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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. 4.11., It is clear from Arrhenius equation (4.18), 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. 4.11: Effect of catalyst on activation energy, not alter Gibbs energy, ΔG 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., , 4.6 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, (4.23), Rate = Z AB e− Ea / RT, 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 (4.23) with Arrhenius equation, we can, say that A is related to collision frequency., Equation (4.23) 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., , * Threshold energy = Activation Energy + energy possessed by reacting species., , 115 Chemical Kinetics

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For example, formation of, methanol, from, bromoethane, depends upon the orientation of, reactant molecules as shown in, Fig. 4.12. The proper orientation of, reactant molecules lead to bond, formation, whereas, improper, orientation makes them simply, bounce back and no products are, formed., Fig. 4.12: 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.,, , Rate = PZ ABe − Ea / RT, 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., , Intext Questions, 4.7, 4.8, 4.9, , What will be the effect of temperature on rate constant ?, The rate of the chemical reaction doubles for an increase of 10K in absolute, temperature from 298K. Calculate Ea., 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?, , Summary, 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, Chemistry 116

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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 ABe − Ea / RT ., , Exercises, 4.1, , 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+, , 4.2, , 2H2O (l) + I3−, , → CH4 (g) + CO(g), → C2H4 (g) + HCl (g), , (iii) CH3CHO (g), (iv) C2H5Cl (g), , →, , Rate = k[H2O2][I-], , Rate = k [CH3CHO]3/2, Rate = k [C2H5Cl], , 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., , 4.3, , 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?, , 4.4, , 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.,, , Rate = k ( pCH3OCH3 ), , 4.5, , 3/2, , If the pressure is measured in bar and time in minutes, then what are, the units of rate and rate constants?, Mention the factors that affect the rate of a chemical reaction., 117 Chemical Kinetics

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4.6, , 4.7, 4.8, , 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 ?, What is the effect of temperature on the rate constant of a reaction? How can, this temperature effect on rate constant be represented quantitatively?, In a pseudo first order hydrolysis of ester in water, the following results were, obtained:, t/s, –1, , [Ester]/mol L, , 4.9, , 4.10, , 60, , 90, , 0.55, , 0.31, , 0.17, , 0.085, , A/ mol L–1, , 0.20, , –1, , 0.30, , –1 –1, , r0/mol L s, , 5.07 × 10, , 0.20, , 0.40, , 0.10, –5, , 5.07 × 10, , 0.05, –5, , 1.43 × 10–4, , What is the order of the reaction with respect to A and B?, The following results have been obtained during the kinetic studies of the reaction:, 2A + B, , 4.12, , 30, , (i) Calculate the average rate of reaction between the time interval 30 to, 60 seconds., (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester., A reaction is first order in A and second order in B., (i) Write the differential rate equation., (ii) How is the rate affected on increasing the concentration of B three times?, (iii) How is the rate affected when the concentrations of both A and B are, doubled?, 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:, , B/ mol L, , 4.11, , 0, , →, , 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., 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:, , Chemistry 118, , 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

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4.13, , 4.14, , 4.15, , 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, 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., The experimental data for decomposition of N2O5, [2N2O5, , →, , 4NO2 + O2], , in gas phase at 318K are given below:, t/s, , 0, , 2, , 10 × [N2O5]/ 1.63, mol L–1, , 4.16, , 4.17, , 4.18, 4.19, 4.20, , 4.21, , 400, , 800, , 1200 1600 2000 2400 2800 3200, , 1.36, , 1.14, , 0.93, , 0.78, , 0.64, , 0.53, , 0.43, , 0.35, , (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 ?, (v) Calculate the rate constant., (vi) Calculate the half-life period from k and compare it with (ii)., The rate constant for a first order reaction is 60 s–1. How much time will, it take to reduce the initial concentration of the reactant to its 1/16th, value?, During nuclear explosion, one of the products is 90Sr with half-life of, 28.1 years. If 1μg of 90Sr was absorbed in the bones of a newly born, baby instead of calcium, how much of it will remain after 10 years and, 60 years if it is not lost metabolically., For a first order reaction, show that time required for 99% completion, is twice the time required for the completion of 90% of reaction., A first order reaction takes 40 min for 30% decomposition. Calculate t1/2., For the decomposition of azoisopropane to hexane and nitrogen at 543, K, the following data are obtained., t (sec), , P(mm of Hg), , 0, , 35.0, , 360, , 54.0, , 720, , 63.0, , Calculate the rate constant., The following data were obtained during the first order thermal, decomposition of SO2Cl2 at a constant volume., , SO2Cl 2 ( g ) → SO2 ( g ) + Cl 2 ( g ), , Experiment, , Time/s–1, , Total pressure/atm, , 1, , 0, , 0.5, , 2, , 100, , 0.6, , Calculate the rate of the reaction when total pressure is 0.65 atm., , 119 Chemical Kinetics

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4.22, , The rate constant for the decomposition of N2O5 at various temperatures, is given below:, T/°C, 5, , 10 × k/s, , 4.23, , 4.24, , 4.25, , 4.26, , 4.27, , 4.28, , 4.29, , 4.30, , -1, , 0, , 20, , 40, , 60, , 80, , 0.0787, , 1.70, , 25.7, , 178, , 2140, , Draw a graph between ln k and 1/T and calculate the values of A and Ea., Predict the rate constant at 30° and 50°C., The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1, at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value, of pre-exponential factor., Consider a certain reaction A → Products with k = 2.0 × 10 –2s–1. Calculate, the concentration of A remaining after 100 s if the initial concentration, of A is 1.0 mol L–1., Sucrose decomposes in acid solution into glucose and fructose according, to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample, of sucrose remains after 8 hours ?, The decomposition of hydrocarbon follows the equation, k = (4.5 × 1011s–1) e-28000K/T, Calculate Ea., The rate constant for the first order decomposition of H2O2 is given by the, following equation:, log k = 14.34 – 1.25 × 104K/T, Calculate Ea for this reaction and at what temperature will its half-period, be 256 minutes?, The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C, and energy of activation 60 kJ mol–1. At what temperature would k be, 1.5 × 104s–1?, The time required for 10% completion of a first order reaction at 298K is, equal to that required for its 25% completion at 308K. If the value of A is, 4 × 1010s–1. Calculate k at 318K and Ea., The rate of a reaction quadruples when the temperature changes from, 293 K to 313 K. Calculate the energy of activation of the reaction assuming, that it does not change with temperature., Answers to Some Intext Questions, , 4.1, 4.2, 4.3, 4.4, , 4.5, 4.6, 4.8, 4.9, , rav = 6.66 × 10 Ms–1, Rate of reaction = rate of diappearance of A, = 0.005 mol litre–1min–1, Order of the reaction is 2.5, X → Y, Rate = k[X]2, The rate will increase 9 times, t = 444 s, 1.925 × 10–4 s–1, Ea = 26.43 kJ mol–1, 1.462 × 10–19, , Chemistry 120, , –6