Chemical Kinetics and Nuclear Chemistry

CHEMICAL KINETICS AND NUCLEAR CHEMISTRY CLASSIFICATION OF REACTIONS : [IN TERMS OF RATES] (i) There are certain reactions which are too slow e.g. rusting of iron, weathering of rocks. (ii) Instantaneous reactions i.e. too fast e.g. Detonation of explosives, acid-base neutralization, precipitation of AgCl by NaCl and AgNO3. (iii) Neither too fast nor too slow e.g. combination of H2 and Cl2 in presence of light, hydrolysis of ethyl acetate catalysed by acid, decomposition of azomethane RATE OF REACTION : The change in concentration of either reactant or product per unit time. Formula : v = ± dc dt dc = change in concentration in a small interval dt. [–] sign is used when we refer to reactant concentration. [+] sign is used when we refer to product concentration. Ex. N2 + 3H2  2NH3 (i) Rate of formation of ammonia = + (ii) Rate of disappearance of nitrogen = – (iii) Rate of disappearance of hydrogen = – Rate of reaction = + Thus, d[NH 3 ] dt d[N 2 ] dt d[H 2 ] dt d[N 2 ] 1 d[H 2 ] 1 d[NH 3 ] =– =– 2 dt dt 3 dt Rate of reaction = – d[N 2 ] 1 d[NH 3 ] = dt 2 dt or rate of formation of ammonia = Twice the rate of disappearance of nitrogen i.e. d[NH 3 ] 2  – d[H 2 ]  =  dt  dt 3  AVERAGE VELOCITY OF REACTION : Change in the concentration of reactants or products per unit time is called average reaction velocity. If c is the change in the concentration of reactants and product in t time, then Average velocity = ± or c t Average velocity = (–) Change in the concentration of reactants Time Concentration (+) Change in the concentration of products Time Product Reactant Time Unit of average velocity = gram mole Unit of concentration = = gram mole litre–1 second–1 litre×Second Unit of time INSTANTANEOUS RATE OF THE REACTION : is called instantaneous rate. The instantaneous rate of the reaction can be determined by measuring concentration of reactant or product at a instant of time and plotting concentration versus time. The instantaneous rate at any time is determined by the slope of the tangent at a point on the time-concentration curve corresponding to the specified time. Δc The slope of the tangent at a point is the limiting value of . Δt Δc dc Lim = Δt 0 Δt dt In terms of the concentration of reactant, the rate of the reaction = – of reactant decreases with time. Concentration The rate of reaction determined at specified concentration or specified time B Time dc . The –sign indicates that the concentration dt dc . The +sign indicates that the concentration dt of product increases with time. In the reaction if at a time t the concentration of product is x and at time t + dt, the In terms of the concentration of product, the rate of the reaction = + concentration becomes x + dx then the reaction rate = dx . dt For example the rate of reaction : N2 + 3H2 ‡ˆ ˆˆ ˆ† ˆˆ 2NH3 in terms of the concentrations of N2, H2 and NH3 can be expressed as : – d[N2 ] 1 d[H2 ] 1 d[NH3 ] ,– ,+ dt 3 dt 2 dt CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex . Sol. From the concentrations of R at different times given below, calculate the average rate of the reaction: R  P during different intervals of time. t/s 0 5 10 20 30 3 –1 10 × [R]/mol L 160 80 40 10 2.5 We can determine the difference in concentration over different intervals of time and thus determine the rate by dividing [R] by t. [R]1 ×10 3 mol L–1 Ex. Sol. [R]2 ×10 3 mol L–1 t2 s t1 s rav ×103 –[R 2 – R 1 ]×10 3 –1 –1 = [t 2 – t1 ] mol L s 160 80 5 0 16 80 40 10 5 8 40 10 20 10 3 10 2.5 30 20 0.75 What should be (a) the rate of disappearance of B and (b) the rate of formation of C, if the rate of disappearance of A for the reaction A + B  2C is 10–2 mole/litre/second at a particular temperature ? (a) Rate of disappearance of A = Rate of disappearance of B = 10–2 mole/litre/second (b) Rate of disappearance of A = 1 × Rate of formation of C 2 Rate of formation of C = 2 × Rate of disappearance of A –2 = 2 × 10 mole/litre/second Ex. Sol. Ex. Sol. Ex. Sol. A gaseous reaction : 2A(g) + B(g)  2C(g), Show a decrease in pressure from 120 mm to 100 mm in 10 minutes. The rate of appearance of C is [A] 2 mm/min [B] 4 mm/min [C] 10 mm/min [D] 12 mm/min. Ans. [B] Suppose 2p is the pressure of C after 10 min. Fall in pressure of A = 2p ; Fall in pressure of B = p Total fall in pressure = (2p + p) – 2p = p = 20 mm Pressure of C = 2p = 40 mm Rate of appearance of C = 40/10 = 4 mm/min dx in the rate expression refers to the dt [A] concentration of the reactants [B] increase in concentration of the reactants [C] instantaneous rate of the reaction [D] average rate of the reaction It is expression for instantaneous rate The term Ans. [C] Which of the following expression can be used to describe the instantaneous rate of the reaction ? 2A + B A2B 1 d[A] d[A] 1 d[A 2 B] 1 d[A] d[B] . [A] – [B] – [C] [D] – Ans. [A] 2 dt dt 2 dt 2 dt dt The instantaneous rate of the reaction can be expressed by any of the following expressions – d[A 2 B] d[B] 1 d[A] or – or dt dt 2 dt Ex. Which of the following will react at the highest rate ? [A] 1 mol of A and 1 mol of B in a 1 L vessel [B] 2 mol of A and 2 mol of B in a 2 L vessel [C] 3 mol of A and 3 mol of B in a 3 L vessel [D] All would react at the same rate Ans. [D] Sol. Since all have same conc. of reactants, all would react at same rate. FACTORS AFFECTING THE RATE OF REACTION : (i) Concentration : Law of mass action enunciates that greater is the conc. of the reactants, the more rapidly the reaction proceeds. (ii) Pressure (Gaseous reaction) : On increasing the pressure, volume decreases and conc. increases and hence the rate increases. (iii) Temperature : It is generally observed that rise in temperature increases the reaction rate. (iv) Nature of the reactants : The rate depends upon specific bonds involved and hence on the nature of reactants. g>l >s (v) Surface area of the reactants : In heterogeneous reactions, more powdered is the form of reactants, more is the velocity. [as more active centres are provided] (vi) Catalyst : Affects the rate immensely. Ex. For the reaction : 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Given : d[NO] = 3.6 × 10–3 mol L–1 s–1 dt Calculate : (i) rate of disappearance of ammonia (ii) rate of formation of water Sol. From the eqn. it is clear that Rate =– or – 1 d[H2O] 1 d[NH3 ] 1 d[NO] = = 4 4 dt dt 6 dt d[NH3 ] d[NO] = = 3.6 × 10–3 mol L–1 s–1 dt dt Thus : Also – 1 d[NH3 ] 1 d [NO] = 4 4 dt dt 1 d[H2O] 1 d[NO] = 4 dt 6 dt 3 d[NO] d[H2O] = 2 dt dt d[H2O] 3 × 3.6 × 10–3 = 2 dt d[H 2 O] = 5.4 × 10–3 mol L–1 s–1 dt Ex. The following reaction was studied in a closed vessel. ˆ ˆˆ †ˆ 4NO2(g) + O2(g) 2N2O5(g) ‡ It was found that concentration of NO2 increases by 4.0 × 10–2 mol L–1 in five seconds, calculate (a) the rate of reaction (b) the rate of change of concentration N2O5. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Sol. (a) Rate = 1 d[NO 2 ] 4 dt d[NO 2 ] 4.0×10 –2 mol L–1 = = 8 × 10–3 mol L–1 s–1 dt 5sec 1  Rate of reaction = × 8 × 10–3 mol L–1 s–1 = 2 × 10–3 mol L–1 s–1 4 Rate of change of conc. of N2O5 –d[N 2 O5 ] 1 = – × Rate of formation of NO2 = dt 2 But (b) =– Ex. 1 [d[N 2 O5 ]] 1 = – × 8 × 10–3 = 4 × 10–3 mol L–1 s–1 2 dt 2 The rate of change in concentration of R in the reaction, 2P + Q  2R + 3S, was reported as 1.0 mol L–1 sec–1. Calculate the reaction rate as well as rate of change of concentration of P, Q and S. Sol. d[Q] 1 d[R] 1 d[S] 1 d[P]    = Rate of reaction dt 2 dt 3 dt 2 dt  d[R] = 1.0 molL–1 s–1 dt  – d[P] d[R] = = 1.0 mol L–1 s–1 dt dt –d[Q] 1 d[R] 1 = = = 0.5 molL–1 s–1 dt 2 dt 2 –d[S] 3 d[R] 3 = = × 1 = 1.5 molL–1 s–1 dt 2 dt 2 Rate of reaction = 1 d[C] 1 = × 1 = 0.5 mol L–1s–1 2 dt 2 SPECIFIC REACTION RATE : Applying law of mass action to the reaction : m1A + m2B  n1C + n2D Rate  [A]m1 [B]m2 or m2 v = k [A]m1 [B] This equation is known as rate law. Where k is the proportionality constant and is called (i) Velocity constant or (ii) Velocity coefficient or (iii) Specific reaction rate. On putting [A] = [B] = 1, where have : v = k Hence specific reaction rate is the rate of the reaction when the concentration of each reactant is taken as unit. Unit of Specific Reaction Rate m m v = k [A] 1 . [B] 2 conc. m +m = k [conc.] 1 2 time –1 [conc.][1– (m1 +m 2 )] × [time] = k or  mole  k=   litre  [1–(m1 +m 2 )] . [second–1] DISTINCTION BETWEEN UNIT OF RATE AND RATE CONSTANT : Rate of a reaction : Its units are always mole litre–1 time–1. Rate constant : Its unit depends upon the order of reaction. RATE LAW : (a) It may also not depend upon the concentration of each reactant or product of the reaction. Suppose, mA + nB  Product R  [A]m [B]n (b) Rate of a chemical reaction is directly proportional to the product of the concentration of reactants raise to the power of their stoichiometric coefficient. (c) The rate law represents the experimentally observed rate of reaction which depends upon the slowest step of the reaction. (d) Rate law cannot be deduce from the equation for a given reaction. It can be find by experiments only. (e) The rate law may not bear a simple relationship of the stoichiometric equation. Ex. In the reaction, A + 2B  6C + 2D, if the initial rate – – Sol. d[A] at t = 0 is 2.6 × 10–2 M sec–1, what will be the value of dt d[B] at t = 0 ? dt [A] 8.5 × 10–2 M sec–1 [B] 2.5 × 10–2 M sec–1 [C] 5.2 × 10–2 M sec–1 [D] 7.5 × 10–2 M sec–1 Ans. [C] From the reaction it is evident that when a mole of A is reacting, 2 moles of B must react. Hence the decrease in the concentration of B must be twice that of A  – d[B] =2 dt  d[A]   – dt    = 2 × 2.6 × 10–2 = 5.2 × 10–2 M sec–1 Ex. The dimensions of rate constant of a second order reaction involves : [A] time and concentration [B] neither time nor concentration [C] time only [D] concentration only Ans. [A] CHEMICAL KINETICS AND NUCLEAR CHEMISTRY s –1 Rate mol L–1 s –1 = (mol L–1)–1 s–1 –1 2 = 2 = (mol L ) mol L–1 [A] Sol. k= Ex. The rate constant of a reaction has same units as the rate of reaction. The reaction is of [A] zero order [B] first order [C] second order [D] none of these Ans. [A] For a zero order reaction, r = k[A]°. Thus the units of k are the same as that of rate of reaction. Sol. Ex. The rate constant of nth order reaction has units : [A] litre1–nmol1–nsec–1 2 Sol. k= [D] mol1–nlitren–1sec–1 Ans. [D] rate [conc.]n Units of k = Ex. 2 [C] mol1–n litre n sec–1 For an nth order reaction : rate = k [conc.]n [B] mol1–nlitre1–nsec mol L–1 s –1 = mol1–n Ln–1 s–1 (mol L–1 ) – Sol. On which of the following factors, the rate constant does not depend ? [A] Temperature [B] Concentration [C] Presence of catalyst [D] Nature of reactants Rate constant is independent of the conc. of the reactants. Note: ORDER OF REACTION : The sum of the power of the concentration terms on which the rate of reaction actually depends as observed experimentally is called the order of the reaction. For example, Order of reaction = x + y Thus, the order of reaction may also be defined as the sum of the exponents (powers) to which the concentration terms in the rate law equation are raised in order to express the observed rate of the reaction. Thus, reaction is said to be of the first order if its rate is given by the expression of the type, r = k1CA Second order if the rate is given by the expression of the type, r = k2C2A or r = k2CACB third order if the rate is given by the expression of the type r = k3CA3 or r = k3CA2CB or r = k3CACB2 or k3CACBCC and so on For zero order reaction, the rate equation is written as R = k0. It is to be noted that the order of reaction is essentially an experimental quantity. Order may be zero, fractional, integer or negative. Ans. [B] Examples showing different values of order of reactions : Reaction Rate law Order (i) 2N2O5 (g)  4NO2 (g) + O2 (g) R = K [N2O5]1 (ii) 5Br– (aq) + BrO3– (aq) + 6H+ (aq) 3Br2 ()+ 3H2O () R= K [Br–] [BrO3–] [H+]2 1+1+2=4 (iii) H2 (Para)  H2 (ortho) R = K [H2 (Para) ]3/2 3/2 1 (iv) NO2 (g) + CO (g)  NO (g) + CO2 (g) R = K [NO2]2 [CO]º 2+0=2 (v) 2O3 (g)  3O2 (g) R = K [O3]2 [O2]–1 2–1=1 (vi) h H2 + Cl2   2 HCl R = K [H2]0[Cl2]0 0+0=0 The reaction (ii) does not take place in one single step. It is almost impossible for all the 12 molecules of the reactants to be in a state of encounter simultaneously. Such a reaction is called complex reaction and takes places in a sequence of a number of elementary reactions. For an elementary reaction the sum of stoichiometric coefficients = order of the reactions. But for complex reactions order is to be experimentally calculated. Ex The rate of a certain reaction depends on concentration according to the equation : –dC K1C = dt 1+ K 2C . What will be the order of reaction, when concentration (C) is : (a) very-very high, (b) very-very low. Sol. (a) K1 K1C – dC = = 1 1+ K 2C dt + K2 C if C is very-very high then  (b)  1 being small may be neglected. C – dC K1 = dt K 2 = constant, i.e., zero order reaction. If C is very-very low 1 + K2C  K K – dC K1C =  1  × concentration =  K  dt K i.e.,  order reaction. Ex. : Reaction H2 + Cl2  2HCl H2 + Br2  2HBr H2 + I2  2HI Experimental rate equation v=k v = k [H2] [Br2]1/2 v = k [H2] [I2] Examples of fractional order reaction Reaction : CO(g) + Cl2(g)  COCl2(g) v = k [CO]2 [Cl2]1/2, order = 2.5 Reaction : COCl2(g)  CO(g) + Cl2(g) v = k [COCl2]3/2, order = 1.5 Ex. For the chemical reaction, 4HBr + O2  2H2O + 2Br2 Rate = k [HBr] [O2] What is the probable mechanism of the reaction ? order zero one and half two CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Sol. Ex. HBr + O2  HOOBr (slow) HOOBr + HBr  2HOBr (fast) HOBr + HBr  H2O + [Br2] × 2 (fast) Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide : 2NO (g) + O2 (g)  2NO2(g) What is the predicted rate law, if the mechanism is :k NO (g) + O2 (g) Sol. NO3 (g) + NO (g) k From the slow step, Rate = k1 [NO3] [NO] From fast step : 1 NO3 (g) (fast) 2NO2 (g) (slow) Equilibrium constant (K) = ..........(i) [NO 3 ] [NO][O 2 ] ..........(ii) Substituting the value of [NO3] from equation (ii) into equation (i), we get Rate= k'[NO]2 [O2] MOLECULARITY OF A REACTION : Note : ‘‘Molecularity is defined as the number of molecules, atoms, or radicals that must collide simultaneously in order for the reaction to take place.’’ It is always a whole number and cannot be negative. In the elementary processes : Participating species Molecularity One species participates ..... unimolecular,1 Two species participates ..... bimolecular, 2 Three species participates ..... trimolecular, 3 Ex. N2O4  2NO2 ...... unimolecular H2 + I2  2HI ...... bimolecular 2FeCl3 + SnCl2  2FeCl2 + SnCl4 ...... trimolecular If the reaction takes place in two or more steps then the overall molecularity of the reaction is monitored by the slow or rate determining step. DIFFERENCE BETWEEN MOLECULARITYAND ORDER OF REACTION : S.No. 1 2 Molecularity Molecularity can neither be zero nor fractional It is the number of molecules of reactants concentration terms taking part in elementary step of a reaction. It can not have negative value. Order of Reaction Order of a reaction can be zero, fractional or integer It is sum of power raised or the rate expression. 4 Molecularity is a theoretical property. Order of a reaction may have negative value. Order is an experimental property. 5 Molecularity concerns with mechanism. Order concerns with kinetic (rate law). 3 Ex. Reaction Rate law Order CH3CHO  CH4 + CO Rate[CH3CHO]3/2 1.5 Rate [NH3]0 0 NH3  1 3 N + H 2 2 2 2 Rate [HI]0, 2Hl  H2 + I2 i.e. Rate = k 0 Order may change with change in experimental conditions while molecularity can't. Ex. isomerization CH3 H2C This reaction follows first order kinetics at high pressure and 2nd order kinetics at low pressure of cyclopropane. PSEUDO UNIMOLECULAR REACTION Consider the reaction : CH3COOC2H5 + H2O H CH3COOH + C2H5OH Since water is present in large excess, its concentration hardly changes during the course of the reaction. And as such rate depends only on the concentration of ester. The order is one but the molecularity is two. Such reactions + Ex. Ans. are called pseudo unimolecular reaction. For a chemical reaction, A  products, the rate of reaction doubles when the concentration of A is increased by 4 times. The order of reaction is [A] 4 [B] 0 [C] 1/2 [D] 1 Ans. [C] r = k[A]n ........ (i) ; 2r = k [4A]n ........... (ii) 2r k  4A  = r k  A  Dividing (ii) by (i) n Sol. or 2 = 22n or 2n = 1 or n = 1/2 For a hypothetical reaction A + B  products, the rate law is, r = k [B] [A]°, the order of reaction is : [A] 0 [B] 1 [C] 2 [D] 3 1+0=1 Ex. The slowest step of a particular reaction is found to be Ex Ans. [B] 1 X + Y2  XY2 2 2 Sol. The order of the reaction is [A] 2 [B] 3 [C] 3.5 r = k[X2]1/2 [Y2]1  Order = 0.5 + 1 = 1.5 Ex. The rate of certain hypothetical reaction A + B + C  products, is given by r= Order of reaction = Ans. [D] dA = k [A]1/2 [B]1/3 [C]1/4 dt The order of a reaction is given by [A] 1 [B] 1/2 Sol. [D] 1.5 [C] 2 1 1 1 6 + 4 + 3 13 + + = = 2 3 4 12 12 [D] 13/12 Ans. [D] CHEMICAL KINETICS AND NUCLEAR CHEMISTRY ZERO ORDER REACTION (A) Reaction whose rate is not affected by concentration said to be of zero order reaction. Ex. : (i) Reaction between Acetone and Bromine (ii) Dissociation of HI on gold surface Unit of Rate Constant : k = mol L–1 sec –1 (B) Unit of rate of reaction = Unit of rate constant. Rate Constant of Zero Order Reaction : x = kt (C) The rate of reaction is independent of the concentration of the reaction substance. Determination of Half life Period of Zero Order Reaction : At t = t½ t1/2 = ; a 2k x= a 2 t1/ 2  a or Ex. The half life period is directly proportional to the initial concentration of the reactants. The rate equation of a reaction is k[A]1/2[B]1/2[C]–1. What should be the order of the reaction ? Sol. n= 1 1 + –1 =0 2 2  Order of the reaction is zero. Graphical representation x 0 Slope = k0 t dx dt 0 t Ex. :  Photochemical reactions, like H2 + Cl2  2HCl, are zero order reaction.  Decomposition of NH3 on platinum surface is also zero order reaction. FIRST ORDER REACTION : When the rate of reaction depends only on one concentration term of reactant. A first order reaction is one whose rate varies as first power of the concentration of the reactant, i.e. the rate increases as number of times as the concentration of reactant is increased. Let us, consider a unimolecular first order reaction represented by the general equation, At t =0 a At t = t a – x A  Product 0 x The initial concentration of A is a mole L–1 and its concentration after time t is (a – x) mole L–1. This means during the time interval t, x mole L–1 of A has reacted. The rate of reaction at any time t is given by the following first order kinetics. d(a – x)  (a – x) dt d(x)  (a – x) dt or or dx = k (a – x) dt where k is the rate constant of the reaction. dx = kdt a–x This is differential rate equation and can be solved by integration. dx  a – x = k  dt or – ln (a – x) = k.t + C ...............(1) where C is integration constant. The constant C can be evaluated by applying the initial condition of the reaction i.e. when t = 0, x = 0. Putting these in equation (1), we get C = – ln a Putting the value of C in equation (1), we get – ln (a – x) = k.t – lna or 2.303 1 a a k = ln = log t a–x t a–x Also, k = and k = ................(2) (a – x1 ) 2.303 log (a – x 2 ) (t 2 – t1 ) 2.303 R log 1 (t 2 – t1 ) R2 where (a – x1) is concentration at time t1 and (a – x2) is concentration after time t2 and R1 is rate at time t1 and R2 is rate at time t2. If [A]0 and [A] be the concentrations of reactant at zero time and time t respectively, then Eq. (2) may be put as 1 [A]0 k = ln t [A] Also, [A] = [A]0 e–kt This is the integrated rate expression for first order reaction. As, k = kt 1 a ln  log(a – x) = log (a) – 2.303 t a–x kt = lna – ln (a – x) Also, (a – x) = ae–kt x = a(1 – e–kt) x  Degree of dissociation   = (1 – e–kt) a CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Unit of Rate constant The differential rate expression for nth order reaction is as follows : dx = k(a – x)n dt – or k= dx (concentration) = = (conc.)1–n time–1 (a – x) n dt (concentration) n time If concentration be expressed in mole L–1 and time in minutes, then k = (mole L–1)1–n min–1 For zero order reaction, n = 0 and hence, k = mole L–1 min–1 For first order reaction, n = 1 and hence, k = (mole L–1)0 min–1 = min–1 For second order reaction, n = 2 and hence, k = (mole L–1)–1 min–1 = mole–1 L min–1 The rate constant of a first order reaction has only time in its unit. It has no concentration term in the unit. This means the numerical value of k for a first order reaction is independent of the unit in which concentration is expressed. If concentration unit is changed, the numerical value of k for a first order reaction will not change. However, it would change with change in time. Say, k is 6.0 × 10–3 min then it may also be written as 1 × 10–4 s–1, i.e., numerical value of k will decrease 60 times if time unit is changed from hour to minute or from minute to second. Half - time or half - life period of a first order reaction : The half - time of a reaction is defined as the time required to reduce the concentration of the reactant to half of its initial value. It is denoted by the symbol t1/2. Thus, When x = a , t = t1/2 2 Putting these values in Eq. (2), we get k= 2.303 log t1/2 a a a– 2 = 2.303 2.303 log 2 = × 0.30103 t1/2 t1/2 ( log 2 = 0.30103) t1/2 = 0.693 k ..............(3) Since k is a constant for a given reaction at a given temperature and the expression lacks any concentration term so from Eq. (3) it is evident that half-time of a first order reaction is a constant independent of initial concentration of reactant. This means if we start with 4 moles L–1 of a reactant reacting by first order kinetics, then after 20 minutes it is reduced to 2 moles L–1. That is, after 20 minutes from the start of reaction the concentration of the reactant will be 2 moles L–1 after 40 minutes from the start of reaction, the concentration is 1 mole L–1. After 60 minutes from the start of reaction, the concentration of the reactant will be reduced to 0.5 mol L–1. In other words, if during 20 minutes 50% of the reaction completes, then in 40 minutes 75%, in 60 minutes 85.5% of the reaction and so on, will complete as shown in the figure above. 1 Thus, fraction left after n half-lives =   2 n n 1 Concentration left after n half-lives, [A] =   [A]0 2 It is also to be noted that Eq. (3) helps to calculate t1/2 or k. A general expression for t1/2 is as follows. Half - life of a nth order reaction Let us find out t1/2 for nth order reaction where n  1.  [A]0 –d[A] –d[A] =kn[A]n  = kdt dt [A]n [A]0  [A]0 [A] d[A] = knt1/2 –n 2 –  2 [A]0 t 1/2 d[A] = k  dt n [A] 0 [A]0  [A]1–n  = knt1/2  1 – n  [A]0/2  1– n  [A]0 1–n    A 0 –   = knt1/2  2     1 1 n  1 n–1 ] = knt1/2 n –1 [1 – 2 (1 – n)  A 0   1–n 1 n [A]0–   1   1 –    = knt1/2 1– n   2   2n –1 – 1 k n (n – 1)[A]0 = t1/2 (order n  1) Therefore, for nth order reaction, the half-life is inversely related to the initial concentration raised to the power of (n – 1). 1 a n –1 where n = order of reaction. t1/2  Ex. (A) (i) All radioactive reactions (ii) (iii) A  Product 2NO ‡ˆ ˆˆ ˆ† ˆˆ 2N2 + O2 (iv) 2Cl2O7  2Cl2 + 7O2 Unit of rate constant of first order reaction K = (sec)–1 (B) Velocity constant for first order reaction k1 = (C) n = 1 2.303 a log10 t (a – x)  t= 2.303 a log10 k1 (a – x) where t = time, a = initial concentration at t = 0 (a – x) = concentration after time t K = Rate constant Graphical Representation Graph between t v/s log a is a straight line (a – x) t Slope = log a/(a–x) 2.303 k CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex. Sol. t 0.75 Calculate t for a 1st order reaction : 0.50 2.303 C0 C0 2.303 k= t log 1  log C0 t1/2 3/4 C0 4 2 log 4 2log 2 t 3/4 = = log 2 log 2 = 2 t1/2  Ex. At least how many half-lives should elapse for a 1 st order reaction A  products so that the reaction is at least 95% completed ? (log 2 = 0.3) (A) 4 (B) 5 (C) 6 (D) 7 Sol. (B) Ex. Sol. A first order reaction gets 90% completed in 40 minute. Find out the half-life period of the reaction. Suppose that the initial concentration of reactant (a) = 100 t = 40 minutes 90% of the reaction get completed in 40 minutes. Therefore, x = 90 100 0% k1 = t1/2 50 50% t1/2 25 75% t1/2 12.5 87.5% t1/2 6.25 93.75% t1/2 3.125 96.875% a 2.303 log a–x t 2.303 100 = 2.303 log  k1 = log 10 40 100 – 90 40 = 2.303 × 1 = 5.757 × 10–2 minutes–1 40 t½ = Ex. Sol. 0.693 = 10.3 minutes 5.757 × 10 –2 Prove with the help of the following data that hydrolysis of H2O2 is a first order reaction. Initial concentration in the reaction 25.0. Time, t 10 20 30 (in minutes) V 20.0 15.7 12.5 For a first order reaction, k1 = 2.303 a log t a–x ; Here, a = 25 k1 at t = 10 minutes = 25 2.303 2.303 log = × 0.0969 = 2.23 × 10–2 20 10 10 k1 at t = 20 minutes = 2.303 2.303 25 log = × 0.2020 = 2.32 × 10–2 20 20 15.7 k1 at t = 30 minutes = 2.303 25 2.303 log = × 0.3010 = 2.31 × 10–2 30 12.5 30 Constant value of k shows that hydrolysis of H2O2 in aqueous medium is a first order reaction. Ex. The rate constant for an isomerization reaction, A  B, is 2.5 × 10–3 min–1. If the initial concentration of A is 1M, calculate the rate of reaction after 1 hr. Sol. As, k =  a , 2.303 log for first order reaction (a – x) t 2.5 × 10–3 = 1 2.303 log (a – x) 60  (a – x) = 0.8607 The rate after 60 minutes = k (a – x) = 2.5 × 10–3 × 0.8607 = 2.1518 × 10–3 Ex. Sol. The rate of a first order reaction is 0.08 mol L–1 at 20 min. and 0.06 mol L–1 at 40 min. after start of reaction. Find the half-life of reaction. As, rate = k[A] 0.08 = k[A]10 0.06 = k[A]20  [A]10 0.08 4 = = [A]20 0.06 3 For first order reaction : [A]10 [A]20 t = 2.303 log when t = (40 – 20) = 20 min. 2.303 4  20 = log k 3 2.303 4 log = 0.0144 min–1 20 3  k=  t1/2 = 0.693 0.693 = = 48.13 min k 0.0144 Ex. Radioactive decay of an atomic nucleus is a first order reaction. Half-life period of radium [ 88 Ra226] is 1590 years. Find out its decay constant. Sol. t½ = 0.693 0.693 0.693 = = = 4.358 × 10–4 Y–1 t1/2 1590 k1 SECOND ORDER REACTION At t = 0 a At t = t (a – x) At t = t1 (a – x1) At t = t2 (a – x2) As per rate law,  A + A  Product A + B a (a – x) (a – x1) (a – x2)  Product 0 x x1 x2 dx = k2[A]n = k2[A]2 = k2 [A][B] dt  dx  2   = k2 (a – x)  dt  (k2 = rate constant for second order reaction) CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Also, 1 1 1 1 a – = k2 =  t  (a – x) a  t a(a – x) or k2= 1  1 1  – (t 2 – t1 )  (a – x 2 ) (a – x1 )  Where (a – x1) and (a - x2) are the concentration of the reactant A at time t1 and t2 respectively. If reactant A and B have different concentrations a and b, then k2 = when a >> b then (a – b) a (a – x)  a Equation reduces to k2 = 2.303 b(a – x) log10 t(a – b) a(b – x) 2.303 b log10 t×a a–x k2' = k2 × a =  2.303  b  log10   b–x  t (equation for first order kinetics) This is an example of pseudo first order reaction. Equation for second order reaction can be rewritten as a 1 = k2t + (a – x) a Graphical Representation 1 a–x In general for nth order reaction, (i) Slope = k2 (ii) Intercept, OA = A 0 1 a t 1  1 1  kn = (n –1)t  (a – x) n–1 – a n–1    In general for nth order reaction, t1/2(n) = 2n–1 –1 (n  2) (n –1)k n (a)n–1 1  t1/2(n)   t1/2(n)  a(1–n) a (n–1) Ex. Initial concentrations of both the reactants of a second order reaction are equal and 60% of the reaction gets completed in 3000 seconds. How much time will be taken in 20% completion of the reaction ? Sol. k2 = 1 x t a(a – x) Suppose, a = 1 Now, for 20% completion x 1 k2 = a(1 – x) t k2 = 0.6 1 1 0.6 × = × 3000 1(1 – 0.6) 3000 0.4 (0.2) 1 0.6 1 × = × 1(1 – 0.2) 3000 0.4 t 1 0.6 1 1 × = × 3000 0.4 t 4 t= 3000 0.4 × 0.6 4 t = 500 second Ex. A second order reaction requires 70 minutes to change the concentration of reactants from 0.08 M to 0.01 M. How much time will it require to become 0.04 M. Sol. For second order reaction when, (a – x) = 0.01 x t.a(a – x) 0.07 k2 = 70 × 0.08 × (0.01) k2 = ......(1) (a – x) = 0.04 k2 = 0.04 t × 0.08× (0.04) .......(2) From the equation (1) and (2) 0.04 0.07 = 70 × 0.08 × (0.01) t × 0.08 × (0.04) t = 10 minutes Ex. Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below : t/min 0 30 60 90 c/M 0.8500 0.8004 0.7538 0.7096. Show that it follows a pseudo first order reaction as the concentration of H 2O remains nearly constant (51.2 M) during the course of the reaction. What is the value of k in the equation? rate = k [CH3COOCH3] [H2O] Sol. For pseudo first order reaction, the reaction should be first order with respect to the ester when [H2O] = constant. From the above data we note t c k{H2O] = ln (c0 / c) min–1 t 0 0.8500 – 30 0.8004 2.004 × 10–3 60 0.7538 2.002 × 10–3 90 0.7096 2.005 × 10–3 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 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 [51.2 M] = 2.004 × 10–3 min–1 k = 3.914 × 10–5 M–1 min–1 It has the units of a second order reaction. THIRD ORDER REACTION A reaction is said to be of third order if its rate is determined by the variation of three concentration terms. When the concentration of all the three reactants is same or three molecules of the same reactant are involved, the rate expression is given as (i) 2NO + O2 ‡ˆ ˆˆ ˆ† ˆˆ 2NO2 (ii) A + B + C  Product Rate constant of third order reaction 1 x(2a – x) k3 = . 2 t 2a (a – x) 2 t1/2 = Half life period 3 2ka 2 Thus, half life is inversely proportional to the square of initial concentration. nth order reaction : A  Product  1 1  – n –1   n –1 (a – x) a   knt = 1 n –1 t1/2 =  2 n –1 – 1  1 .  n –1  k n (n – 1)  a  [n  1, n = order] Side or concurrent reaction : B k1 A ; k2 ln C [A]0 = (k1 + k2) t [A]t ; [B] k1 = [C] k 2 Consecutive reaction : A k1 B k2 C : k 1 tmax = ln 1 k1 – k 2  k 2 k2   k 2  k1 –k 2  : [B]max = [A]0     k1  Graphical comparison of different orders 1 st order t t Conc. 2 nd order Rate (Conc.)2 Conc. 2 nd order 1 st order Zero order t 1/2 t 1/2 t 1/2 t 1 st order Rate Rate Zero order Conc. 2 nd order 1 [ A] Log [A] Conc.[A] Zero order Conc. 1/conc. THRESHOLD ENERGY AND ACTIVATION ENERGY : For a reaction to take place the reacting molecules must colloid together, but only those collisions, in which colliding molecules possess certain minimum energy is called threshold energy (ET). Activation energy (Ea) : The extra energy needed for the reactant molecules to be able to react chemically is known as Activation energy. ET = Threshold energy Ea = Activation energy of forward reaction E'a = P1 = P2 = activation energy of backward reaction Potential energy of reactants Potential energy of products EFFECT OF CATALYST A catalyst is a substance, which increases the rate of reaction without itself being consumed at the end of the reaction, and the phenomenon is called catalysis. There are some catalysts which decrease the rate of reaction and such catalysts are called negative catalyst. Obviously, the catalyst accelerating the rate will be positive catalyst. However, the term positive is seldom used and catalyst itself implies positive catalyst. Catalysts are generally foreign substances but sometimes one of the product formed may act as a catalyst and CHEMICAL KINETICS AND NUCLEAR CHEMISTRY such a catalyst is called "auto catalyst" and the phenomenon is called auto catalysis. Thermal decomposition of KClO3 is found to be accelerated by the presence of MnO2. Here, MnO2 (foreign substance) act as a catalyst 2KClO3 + [MnO2]  2KCl + 3O2+ [MnO2] MnO2 can be received in the same composition and mass at the end of the reaction. In the permanganate titration of oxalic acid in the presence of bench H2SO4 (acid medium), it is found that there is slow discharge of the colour of permanganate solution in the beginning but after sometime the discharge of the colour becomes faster. This is due to the formation of MnSO4 during the reaction which acts as a catalyst for the same reaction. Thus, MnSO4 is an "auto catalyst" for this reaction. This is an example of auto catalyst. 2KMnO4 + 4H2SO4 + 5H2C2O2  K2SO4 + 8H2O + 10CO2 GENERAL CHARACTERISTIC OF CATALYST  A catalyst does not initiate the reaction, it simply fastens it.  Only a small amount of catalyst can catalyse the reaction.  A catalyst does not alter the position of equilibrium i.e. magnitude of equilibrium constant and hence G°. It simply lowers the time needed to attain equilibrium. This mean if a reversible reaction in absence of catalyst completes to go to the extent of 75% till attainment of equilibrium, and this state of equilibrium is attained in 20 minutes then in presence of a catalyst also the reaction will go to 75% of completion before the attainment of equilibrium but the time needed for this will be less than 20 minutes. A catalyst drives the reaction through a different route for which energy barrier is of shortest height and Hence,  Ea is of lower magnitude. That is, the function of the catalyst is to lower down the activation energy. P.E. Ea = Energy of activation in absence of catalyst. Ea E'a HR E'a = Energy of activation in presence of catalyst. Reactants HP Products Ea – E'a = Lowering of activation energy by catalyst. Reaction Coordinate If k and kcat be the rate constant of a reaction at a given temperature T, and Ea and E'a are the activation energies of the reaction in absence and presence of catalyst, respectively, the k cat Ae –E'a /RT = –Ea /RT k Ae k cat = Ae (Ea –E'a )/RT k k cat gives the number of times the rate of reaction will increase by the k use of catalyst at a given temperature and this depends upon E a – E' a. Greater the value of Ea – E'a, more number of times kcat is greater than k. The rate of reaction in the presence of catalyst at any temperature T1 may be made equal to the rate of reaction in absence of catalyst but we will have to raise the temperature. Let, this temperature be T 2, then Since Ea, E'a is +ve, so kcat > k. The ratio e –E'a /RT1 = e –Ea /RT2 or E'a E a = T1 T2 Ex. A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in presence of catalyst at the same rate, the temperature required is 400K. Calculate the activation energy of the reaction if the catalyst lowers the activation energy barrier by 40 kJ/mol. Sol. Let, Ea and E'a be the energy of activation in absence and presence of catalyst for hydrogenation reaction, as k  Ae  Ea / RT k1  Ae  Ea / R500 (In absence of catalyst) k 2  Ae  E 'a / R  400 (In presence of catalyst) Given, r1 = r2 ; Hence k1 = k2 e  Ea / R  5 00  e  E 'a / R  400  Ea E'a = R ×500 R ×400  Ea = 200 kJ/mol–1 or E a E a – 40 = 500 400 (As Ea – E'a = 40) DETERMINATION OF ORDER OF REACTION Integration Method In this method, value of k is determined by putting values of initial concentration of reactants and change in concentration with time in kinetic equation of first, second and third order reactions. The equation by which constant value of k is obtained is called order of that reaction. Ex. a 2.303 log a–x t k2 = 1 t k3 = 1  x(2a – x)   2 2 2t  a (a – x)  (For first order reaction)  x   a(a – x)    (For second order reaction) (For third order reaction) For a reaction, A  B, it has been found that the order of the reaction is zero with respect to A. Which of the following expression correctly describes the reaction ? [A] k = Sol. k1 = – [A] 2.303 log 0 t [A] [B] [A]0 – [A] = kt [C] t½ = 0.693 k [D] t½  1 [A]0 Ans. [B] d[A] = k[A]0, – d[A] = kdt dt Integrating from t = 0 to t = t [A]0 – [A] = kt Graphical Method If a straight line is obtained on drawing a graph between log (a – x) and time then it is first order reaction. If a straight line is obtained on drawing a graph between (a – x)2 and dx , then it is second order reaction. dt CHEMICAL KINETICS AND NUCLEAR CHEMISTRY [A] [B] Rate Rate Rate Which of the following graphs is for a second order reaction ? Rate Ex. [C] [D] [A] 2 [A] 2 [A] 2 [A] Ans. [C] Sol. For second order reaction rate vs [A]2 is a straight line with slope equal to k rate = k[A]2 If a straight line is obtained on drawing a graph between (a – x)3 and dx , then it is third order reaction. dt Half-life Method Relation between half-life period of a reaction and initial concentration is as follows : t1/2  Ex. For first order reaction (Half life  a0) For second order reaction (Half life  1/a) For third order reaction (Half life  1/a2) 1 a n –1 For a first order reaction, t0.75 is 1386 seconds. Therefore, the specific rate constant is [A] 10–1 s–1 [B] 10–3 s–1 Sol. t0.75 = 1386 s = 2 × t0.5 ; Ex. t½ of first order reactions is given by [A] 0.693 k [C] 10–2 s–1 t0.5 = 1386 = 693 s 2 [B] ; k= [D] 10–4 s–1 Ans. [B] 0.693 = 1 × 10–3 s–1 693s 0.693 , t3/4 would be equal to k 0.346 k [C] 1.386 k [D] 0.924 k Ans. [C] 2 × 0.693 1.386 = k k Sol. t3/4 = 2(t½ ) = Ex. The t ½ of a first order reaction is found to be 2 minutes. The percentage of the reactant left after 360 seconds is : [A] 12.5 Sol. [B] 25 [C] 15 360 seconds = 6 min = 3 half-lives 100 t1/2 50 t1/2 25 t1/2 12.5 [D] 7.5 Ans. [A] Ex. Sol. The rate of decomposition of N2O5 in CCl4 solution has been studied at 318 K and the following results have been obtained : t/min 0 135 342 683 1693 c/M 2.08 1.91 1.67 1.35 0.57 Find the order of the reaction and calculate its rate constant. What is the half-life period? It can be shown that these data will not satisfy the integrated rate law of zero order. We now try integrated first order equation i.e., k= ln (c 0 / c) t t/min c/M 0 135 342 683 1693 2.08 1.91 1.67 1.35 0.57 ln (c0 / c) min–1 t 6.32 × 10–4 6.30 × 10–4 6.32 × 10–4 6.32 × 10–4 6.31 × 10–4 k= It can be seen that the value of k is almost constant for all the experimental results and hence it is first order reaction with k = 6.31 × 10–4 min–1 . t1/2 = 0.69 = 1.094 × 103 min–1 6.31×10 –4 min –1 Graphical method : Alternatively, if we draw a graph between ln c against t, we obtain a straight line with slope = – k. Ex. In the reduction of nitric gas with hydrogen, the reaction was found to be 50% complete in 210 seconds when the initial pressure of the mixture was 200 mm. In a second experiment the time of half reaction was 140 seconds when the initial pressure was 300 mm. Calculate the total order of the reaction. Sol. For a nth order reaction (n  1), t1/2  210  300  =   140  200  1 c0n–1 n –1 n=2 OSTWALD ISOLATION METHOD This method is used to find out the order of complex reactions. If nA, nB and nC molecules of substance A, B and C, respectively, are present in a reaction, then nA + nB + nC will be the order of reaction. When B and C are in excess, the order of reaction will be nA. When A and B are in excess, the order of reaction will be nC. When A and C are in excess, the order of reaction will be nB. Ex. Sol. When the initial concentration of a reaction was doubled, its half life become half, What should be the order of the reaction ? Rate law for product of a reaction is as follows : Rate = k[A]n and t 0.5  2a  = t 0.5  a  2 n–1 ; 2 = [2] n=2 n–1=2 n –1 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex. CH3COF + H2O  CH3COOH + HF The reaction : has been studied under the following initial conditions : Case I Case II c H0 2O = 1.00 M c H0 2O = 0.02 M c 0CH3COF = 0.01 M c 0CH3COF = 0.80 M Concentration were monitored as a function of time and are given below : Case I Case II t/min c CH COF / M 3 t/min c H2O /M 0 0.01000 0 0.0200 10 0.00857 10 0.0176 20 0.00735 20 0.0156 40 0.00540 40 0.0122 Determine the order of the reaction and the rate constant for the reaction. Sol. a b Let rate = k (cCH3COF ) (c H 2O ) 0 0 We shall use here Ostwald isolation method, as in one set of experimental conditions, c H 2O >> c CH3COF and in the 0 0 second case, c H2O  c CH3COF . In the first case we determine the order of the reaction with respect to CH3COF. Wee note that the reaction is not of zero order as rate of reaction changes with time. We now, apply first order equation and find, t/min cCH3COF / M 0 10 20 40 0.01000 0.00857 0.00735 0.00540  c0CH COF 3 ln   b c k(c H 2O )  CH3COF = –1 t min     – 0.0154 0.0154 0.0154 Therefore, k (c H2O ) = 0.0154 min–1 and we know that the order of reaction with respect to CH3COF is 1. Now we determine the order of reaction with respect to water. Again we try for first order equation. t/min c H 2O / M 0 0.0200 10 0.0176 20 0.0156 40 0.0122 average = 0.0125 min–1  c0H O  2 ln  cH O k(cCH3COF )b  2 = –1 min t – 0.0128 0.0124 0.0124     The reaction is first order in H2O and we have k(cCH3COF ) = 0.0125 min–1 Niw in case , k = and in case  = 0.0154min –1 = 0.0.154 M–1 min–1 1.00 0.0125 = 0.0156 M–1 min–1. 0.800 TEMPERATURE EFFECT The rate of reaction is dependent on temperature. This is expressed in terms of temperature coefficient which is a ratio of two rate constants differing by a temperature of 10°. Generally the temperature selected are 298K and 308K. It is mathematically expressed as, Temperature coefficient = k t + 10 rate constant at 308K = k rate constant at 298K t The value of temperature coefficient for most of the reactions lies between 2 to 3. Ex. Sol. For a reaction T.C. = 2, Calculate k 40ºC for this reaction. k 25ºC Δt 15 3 k2 = (T.C.) 10 = (2)10 = (2) 2 = 8 k1 But the method of temperature coefficient was not exact and to explain the effect of temperature on reaction rate new theory was evolved ARRHENIUS EQUATION Arrhenius derived a mathematical expression to give a quantitative relationship between rate constant and temperature. The expression is k = A.e—Ea /RT (Here, A = frequency factor; Ea = activation energy ; R = gas constant and T = temperature). If k! and k2 are rate constants at temperature T1 and T2 then log Ex.  T2 – T1  k2 Ea =   k1 2.303 R  T1T2  Ethylene oxide is decomposed into CH4 and CO. Rate constant for this reaction may be described by the equation log k (s–1) = 14.34 – (i) (ii) (iii) Sol. 1.25×10 4 T What will be the energy of activation of this reaction ? What will be the value of k at 670 K ? At what temperature will its half-life period be 25.6 minutes ? (i) We know, log10k = log10A – Given, log k (s–1) = 14.34 – Ea 2.303RT 1.25×10 4 T .......(i) .......(ii) CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Comparing Eqs. (i) and (ii), we get Ea = 1.25 × 104 2.303R (ii) Ea = 1.25 × 104 × 2.303 × 8.314 × 10–3  Ea = 239.339 kJ/mol Substituting the value of T (670 K) in Eq. (ii), log k (s–1) = 14.34 –  k = 4.82 × 10–5 s–1 (iii) k= 1.25×10 4 = 4.3167 670 0.693 0.693 = = 0.000451 sec–1 t1/2 256 × 60 log 0.000451 sec–1 Ex. 1.25 ×104  T = 706.79 K T Two st order rxns. are initially having equal rate at a particular temprature. Temprature of both the reaction is increased by same amount. Calculate rate of which reaction will increase by greater amount ( reaction with low E a or high Ea) Sol.  k2  E  1 1 n  k  = a  T – T  R  1 2  1 .....(i)  k 2'  E '  1 1  n  k  = a  T – T '  R  1 2   1 .....(ii) = 14.34 – Equation (i) – (ii).  k 2  ΔT n  k '  = ( Ea – Ea,) R  2 if or k=Ae Ea > Ea, k2 > k2’ E a / RT dk – EaA  –1  –E a /RT  2 = dT RT  T  A e temprature coeff. of rate constant Ea 1 dk = K dT RT 2 Δk k.ΔT = fractional change / unit temp. rise Ex. Explain on the basis of temprature coeff. of rate const. that equilibrium of endothermic reaction shifts in forward direction on increasing temprature while equilibriums of exothermic shift back.  K eq1 n  K  eq2  ΔH  1 1  =  –   R  T1 T2   Sol. H = Eaf – Eab > 0 (for endothermic) Eaf > Eab on increasing temp. kf  more than kb equilibrium will shift in forward direction. Ex. The rate constant of forward reaction a reaction increases by 6% when the temperature of the reaction is increased from 300 to 301 K, whereas equilibrium constant increases by 2%. Calculate the activation energy for the forward as well as backward reaction. Sol. According to Arrhenius equation, log E a(f)  T2 – T1  k2 = k1 2.303R  T1T2  If k1 = k at 300 K then at 301 K, k2 = k + k × 6 = 1.06 k 100 E a(f) 1.06 k  301– 300  = k 2.303×8.314  300×301   log10  Ea (f ) = log (1.06) × 2.303 × 8.314 × 300 × 301 = 43753 J/mol = 43.753 kJ/mol According to van't Hoff equation, log If k1 = k at 300 k, k2 = k + log10 k2 ΔH°  T2 – T1  = k1 2.303R  T1T2  2 = 1.02 k 100 1.02 k ΔH°  301– 300  =   k 2.303×8.314  300×301  H° = (log 1.02) × 2.303 × 8.314 × 300 × 301 J/mol = 14869 J mol–1 = 14.87 kJ/mol Thus, reaction is endothermic. For such a reaction, H° = Ea(f) – Ea(b)  Ea(b) = Ea(f) – H° = 43.753 – 14.87 kJ/mol = 28.883 kJ mol–1 Value of rate constant for a first order reaction at 500 K is 1.60 × 10–5 second–1, whereas at 600 K, it is 6.36 × 10–3 second–1. Find out the activation energy of the reaction.  Ex. Sol. –E a  1 – 1  k    log  1  = T T2   k 2  2.303 R  1 log 1.60 × 10–5 – log 6.36 × 10–3 5.241 – 3.8035 = Ea = 1.49 × 105 Ea 1 × 19.15 3000 = 1   1 –E a – 2.303× 8.314  500 600  Ea = 2.5999 × 19.15 × 3000 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex. An exothermic reaction, X  Y, has an activation energy 30 kJ mol–1. If energy change (E) during the reaction is – 20kJ, then the activation energy for the reverse reaction is [A] –30kJ [B] 20 kJ [C] 50 kJ [D] 10 kJ Ans. [C] Sol. E = E a (f) – E a (b) ; Ex. An endothermic reaction, A  B, has an activation energy as x kJ mol–1 of A. If energy change of the reaction is y kJ, the activation energy of reverse reaction is [A] – x [B] x – y [C] x + y [D] y – x Ans. [B] Sol. E = E a (f) – Ea (b) ; Ex. Which of the following relations is correct ? – 20 = 30 – E a (b) y = x – E a (b) [A] k = A eEa/RT Sol. Sol. Ex. ; [B] ln k – ln A = Ea RT Ea RT or ln A – ln k = Ea (b) = x – y [C] ln A – ln k = Ea RT [D] ln A – ln k = – Which of the following expression give the effect of temperature on the rate constant ? [A] ln A = RT ln Ea – ln k [B] ln k = ln A – Ea/RT [C] k = AEa/RT [D] None of these Ans. [B] The effect of temperature on rate constant is quantitatively given by Arrhenius equation k = Ae–Ea/RT or ln k = ln A – Ea/RT The plot of log k vs 1 helps to calculate T [B] Rate constant of the reaction [D] Energy of activations as well as the frequency factor According to Arrhenius equation : log k = log A – Plot of log k vs. Ea 1 . 2.303 T 1 is straight line T log k Ea Slope = – 2.303R 1/T Intercept = log A Ex. Ea RT Ea RT [A] Energy of activation [C] Order of the reaction Ans. [D] Sol. E a (b) = 50 kJ Ans. [C] k = Ae–Ea/RT ln k = ln A – Ex. ; The progress of the reaction given below, consider the reaction given below, + CH3COOC2H5 + H2O H CH3COOH + C2H5OH the reaction can be followed by measuring the concentration of acid (HCl acid used as catalyst plus acetic acid formed during the reaction) by means of alkali titration. Calculate the volume of alkali (NaOH) needed for the end point that will increase with time. Sol. + CH3COOC2H5 + H2O H CH3COOH + C2H5OH At t = 0 a excess 0 0 At t = t a – x x x At t =  0 a a If V0, Vt and V are the volumes of NaOH solution needed for the end point of titration of the reaction mixture at zero time, time t and at infinity, i.e. after completion of the reaction the condition being achieved by heating the reaction mixture for some time, then V0  [acid catalyst] Vt [acid catalyst] + x V [acid catalyst] + a  V– Vt  a – x V– V0  a (since concentration of HCl acid acting as catalyst will remain constant). The above reaction which is of first order ( actually pseudo unimolecular) will, therefore, obey following equation. k= Ex. 2.303 V – V0 log  t V – Vt H2O2 (aq.)  H2O + 1 O 2 2 At t = 0 a 0 0 At t = t a – x x x Since H2O2 acts as a reducing agent towards KMnO4, so concentrations of H2O2 at various time intervals may be determined by the titration of the reaction mixture against standard KMnO 4 solution. The titre value will go on decreasing with time. If V0 and Vt be the titre values at zero time and any time t then V0 a and Vt a – x The above reaction being first order, its rate constant may be expressed as k= Ex. V0 2.303 log V t t The reaction mentioned below is first order w.r.t. sucrose and zero order w.r.t. water, since water is in large excess as compared to sucrose. That is, it is an example of pseudo unimolecular reaction. Sucrose, glucose and fructose all are optically active substances. Therefore, the progress of the reaction can be followed by measuring angle of rotations of the reaction mixture at various time intervals. During the reaction, angle of rotation goes on decreasing and after sometime there is reversal of the direction of rotation, i.e. from dextro to laevo and Hence, the reaction is called "inversion of cane sugar" or inversion of sucrose. C12H22O11 + H2O d-Sucrose Initially a H + C6H12O6 + C6H12O6 d-Glucose -Fructose Excess 0 0 After time t a–x Constant x x At infinity 0 Constant a a Angle of optical rotation is measured by means of an instrument called polarimeter. Optical rotation is mathematically expressed as, Robs= .C. []Dt CHEMICAL KINETICS AND NUCLEAR CHEMISTRY where  = length of the polarimeter tube C = concentration of test solution []Dt = specific rotation For a given sample and polarimeter,  and []Dt are constant. Robs C, or Robs = kC, If r0, rt and r be the observed angle of rotations of the sample at zero time, time t and infinity respectively, and k1, k2 and k3 be proportionate in terms of sucrose, glucose and fructose, respectively. Then, r0 = k1a rt = k1(a – x) + k2x + k3x r = k2a + k3a From these equations it can be shown that a r –r = 0  a – x rt – r So, the expression for the rate constant of this reaction in terms of the optical rotational data may be put as k = r0 – r 2.303 log rt – r t Ex. 2N2O5  4NO2 + O2 At t = 0 P0 0 0 At t = t P0 – 2x 4x x At t =  0 2P0 ½ P0 The progress of the reaction can be followed by measuring the pressure of the gaseous mixture in a closed vessel, i.e. at constant volume. The expression for the rate constant in terms of pressure data will be as given below. k= 2.303 P log 0 , where Pt = P0 – 2x t Pt If total pressure after any time t and at  is given, then it is possible to find P0 and x and hence, k may be calculated. Ex. Sol. Consider a first order reaction, A B + C Assume that A, B and C are gases. The given data is Time 0 T Partial pressure of A P1 P2 And we have to find the rate constant of the reaction. Since A is a gas assuming it to be ideal, we can state that PA = [A] RT [From PV = nRT]  At t = 0, P1 = [A]0 RT and at t =t, P2 = [A]tRT. Thus, the ratio of the concentration of A at two different time intervals is equal to the ratio of its partial pressure at those same time intervals.  [A]0 P1 = [A]t P2  k= 1 P1 ln t P2 Ex. Sol. A B + C Time 0 t Total pressure of A + B + C P1 P2 Find k. In this case, we are given total pressure of the system at these time intervals. The total pressure obviously includes the pressure of A, B and C. At t = 0, the system would only have A. Therefore, the total pressure at t = 0 would be the initial pressure of A.  P1 is the initial pressure of A. At time t, let us assume moles of A will decompose to give B and C because of which its pressure is reduced by an amount x while that of B and C is increased by x each. That is : A  B + C Initial P1 0 0 At time t P1 – x x x  Total pressure at time t = P1 + x = P2  x = P2 – P 1 Now the pressure of A at time t would be P1 – x = P1 – (P2 – P1) = 2P1 – P2  Ex. k = ln [A]0 P1 = ln [A]t (2P1 – P2 ) For the given following first order reaction, A  B + C Sol. Time T  Total pressure of A + B + C P2 P3 Calculate k. Here  means that the reaction is complete. Now, we have A At t = 0 P1 At t = t (P1 – x) At t =  0  2P1 = P3  P1 =  B 0 x P1 + C 0 x P1 P3 2 At time t, P1 + x = P 2  P3 P3 + x = P2  x = P2 – 2 2  P1 – x = k= P3 P3 – (P2 – ) = P3 – P2 2 2 1 P3 [A]0 1 P /2 1 = ln 3 ln = ln t 2(P3 – P ) [A]t t (P3 – P2 ) t CHEMICAL KINETICS AND NUCLEAR CHEMISTRY CATALYST AND CATALYSIS : A catalyst is a substance, which increases the rate of a reaction without itself being consumed at the end of the reaction, and the phenomenon is called catalysis. Types of catalyst Negative catalyst Positive Catalyst increases the rate of reaction decreases the rate of reaction Catalyst are generally foreign substances but sometimes one of the product may act as a catalyst and such catalyst is called “auto catalyst” and the phenomena is called auto catalysis. Examples of catalysis (a) Thermal decomposition of KCIO3 is found to be accelerated by the presence of MnO2. Here MnO2 acts as a catalysts. 2KClO3 + [MnO2]  2KCl + 3O2  + [MnO2] MnO2 can be received in the same composition and mass at the end of the reaction. (b) In the permanganate titration of oxalic acid initially there is slow discharge of the colour of permanganate solution but afterwards the discharge of the colour become faster. This is due to the formation of MnSO 4 during the reaction which acts as a catalyst for the same reaction. Thus, MnSO4 is an “auto catalyst” for this reaction. This is an example of auto catalyst. 2KMnO4 + 3H2SO4 + 5H2C2O4  K2SO4 + 8H2O + 10CO2 + 2MnSO4 General Characteristics of Catalyst  A catalyst does not initiate the reaction. It simply fastens it.  Only a small amount of catalyst can catalyse the reaction.  A catalyst does not alter the position of equilibrium i.e. magnitude of equilibrium constant and hence Gº. It simply lowers the time needed to attain equilibrium. This means if a reversible reaction in absence of catalyst completes to go to the extent of 75% till attainment of equilibrium, and this state of equilibrium is attained in 20 minutes then in presence of a catalyst the reaction will go to 75% of completion before the attainment of equilibrium but the time needed for this will be less than 20 minutes.  A catalyst drives the reaction through a low energy path and hence Ea is less. That is, the function of the catalyst is to lower down the activation energy. Ea = Energy of activation in absence of catalyst. E’a = Energy of activation in presence of catalyst. Ea – E’a = lowering of activation energy by catalyst. Comparison of Rates of Reaction in Presence and Absence of Catalyst If k and kcat be the rate constant of a reaction at a given temperature T, and Ea and E’a are the activation energies of the reaction in absence and presence of catalyst, respectively, the k cat Ae –E'a/RT (E a –E'a )/RT = = k Ae –Ea /RT Ae Since Ea – E’a is positive so kcat > k. the ratio use of catalyst at a given temperature k cat gives the number of times the rate of reaction will increase by the k The rate of reaction in the presence of catalyst at any temperature T 1 may be made equal to the rate of reaction in absence of catalyst but for this sake we will have to raise the temperature. Let this temperature be T2 = e–E'a /RT1 = e–Ea /RT2 or E'a E a = T1 T2 Uncatalyzed pathway Ea Reactants Products Potential energy Potential energy Catalyzed pathway Reaction progress (a) Ex. Sol. Ea Reactants Products Reaction progress (b) For the reaction CO(g) + Cl2 (g)  COCl2 (g) under the same concentration conditions of the reactants, the rate of the reaction at 250°C is 1500 times as fast as the same reaction at 150°C. Calculate the activation energy of the reaction. If the frequency factor is 2.0 × 10 10 M–1 sec–1, calculate the rate constant of the reaction at 150°C.  T2 – T1  k2 E log k = 2.303 R  T T  1  1 2  log 1500 = E= E 100 × 2.303× 2 523× 423 3.1761× 2.303× 2×523× 423 = 32.36 kcal mol–1 100 E 32360 log k = log A – 2.303 RT = log (2.0 × 1010) – = 10.301 – 16.609 = – 6.308 2.303× 2× 423 k = 4.92 × 10–7 litres mol–1 sec–1 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex. The pyrolysis of an organic ester follows a first order process and its rate constant can be expressed as 42075 T –1 where k is given in the min . Calculate the time required for 25 percent reaction to complete at 227°C. ln k = 78.09 – Sol. ln k = 78.09 – log k = – 42075 = – 6.06 500 6.06 = – 2.63 ; k = 2.344 × 10–3 min–1 2.303 when x = 0.25 ; k = t1/4 = 2.303 a log t1/4 0.75a 2.303 log 1.333 = 123.06 min 2.344×10 –3 1 for a certain reaction was found to be – 5.4 × 103. Calculate the energy of T activation of the reaction. If the rate constant of the reaction is 1.155 × 10–2 sec–1 at 373 K, what is its frequency factor ? Ex. The slope of the plot of log k vs Sol. –E (a) slope = 2.303 R = – 5.4 × 10 3 E = 5.4 × 103 × 2.303 × 1.987 = 24.624 cal mol–1 (b) E = Ae– E/RT ; log 1.155 × 10 –2 = log A – 24.624 2.303×1.987×373 or A = 1.764 × 103 sec–1 PARALLEL REACTIONS These are reactions in which reaction substances do not follow a particular path to give a particular set of products. It follows one or more paths to give different products, e.g. k1 B A k2 C The reactant A follows two different paths to form B and C as shown below : A C B Conc. t Rate = – d[A] = k1[A] + k2[A] dt = k[A] [As, (k1 + k2) = k] % yield of B = Ex. k1 × 100 k1 + k 2 A follows parallel path, first order reaction giving B and C as k1 B 5A k2 2C If initial concentration of A is 0.25 M, calculate the concentration of C after 5 hours of reaction. [Given : k1 = 1.5 × 10–5 s–1, k2 = 5 × 10–6 s–1] Sol. k= 2.303 [A] log t [A]t (k = k1 + k2 = 1.5 × 10–5 + 5 × 10–6 = 2 × 10–5 s–1) 2.303 0.25 log t [A]t  2 × 10–5 =   [A]t = 0.1744 M [A]decomposed = [A]0 – [A]t = 0.25 – 0.1744 = 0.0756 M Fraction of C formed = k2 2 5×10 –6 2 × [A]decomped × = × 0.0756 × = 7.56 × 10–3 M –5 (k1 + k 2 ) 5 2×10 5 (5 moles of A are used to give 2 moles of C) SEQUENTIAL REACTIONS These are reactions which proceed from reactants to product through one or more intermediate stages, Ex. A k1 B k2 C Graphical representation –d[A] = k1[A] dt .............(1) d[B] = k1[A] – k2[B] dt .............(2) d[C] = k2[B] dt .............(3) A B C Conc. t Integrating Eq. (1), we get [A] = [A]0 e  k1 t Now, we shall integrate Eq. (2) and find the concentration of B related to time t. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY d[B] = k1[A] – k2[B] dt  d[B] + k2[B] = k1[A] dt Substituting [A] as [A]0 e  k1 t  d[B] + k2[B] = k1 [A]0 e  k1 t dt .............(4) Integration of the above equation is not possible as we are not able to separate the two variables, [B] and t. Therefore, k t we multiply Eq. (4) by integrating factor e 2 , on both the sides of the equation.  d[B]  + k 2 [B]  e k 2 t = k1[A]0 e(k 2 –k1 )t   dt  We can see that the left hand side of the equation is a differential of [B] e k2 t .  d ([B]ek 2 t ) = k1[A]0 e(k2 –k1 )t dt d ([B]e k2 t )  k1 [A]0 e (k2 k1 )t dt Integrating within the limits 0 to t.  d ([B]e  k2t t ) = k1[A]0  e(k 2 –k1 )t dt 0 [B]ek 2t = [B] = k1[A]0  (k2 –k1 )t  e –1 (k 2 – k1 ) k1[A]0  –k1t –k 2t  e –e k 2 – k1 t  e(k 2 –k1 )t   [B]ek 2 t = k1[A]0    (k 2 – k1 )  0  [B] = .............(5) Now, in order to find [C], substitute Eq. (5) in Eq. (3), we get d[C] k1k 2 [A]0 –k1t –k 2 t = [e – e ] dt k 2 – k1  d[C] = k1k 2 [A]0  –k1t –k 2t  e –e dt k 2 – k1 On integrating, we get  d[C] = t k1k 2 [A]0 [e –k1t – e –k 2 t ]dt (k 2 – k1 ) 0  e –k1t  t  e –k 2 t  t    –    –k1 0  –k 2  0   [C] = k1k 2 [A]0 (k 2 – k1 )  [C] =  e –k1t –1   e –k 2 t –1   k1k 2 [A]0  –  (k 2 – k1 )  –k1   –k 2   k1[A]0 –k 2t  (k 2 –k1 )t  e e –1 (k 2 – k1 )  [C] =  1– e –k1t   1– e –k 2 t k1k 2 [A]0  – k 2 – k1   k1   k 2 [C] = [A]0 [k 2 (1– e –k1t )(1– e –k 2 t )] k 2 – k1    Bmax and tmax : We can also attempt to find the time when [B] becomes maximum. For this, we differentiate Eq. (5) and find d[B] and equate it to zero. dt  d[B] k1[A]0 e –k1t (–k1 ) + e –k 2 t (k 2 )  = 0 = (k 2 – k1 )  dt  k 1e –k1t = k 2 e –k 2 t k1 k –k t = e 1 2  , taking log of both the sides k2  tmax = 1 k ln 1 k1 – k 2 k 2 ...........(6) Substituting Eq. (6) in Eq. (5)  k2  Bmax= [A]0    k1  k 2 /k1 – k 2 Ex. Starting from the equation for radioactive decay i.e. – Sol. (i) – dN = N . This on integration gives (with N = N0 , at t = 0) ; N = N0 e–t dt    Mean life =     Ex. dN 1 = N, derive the expression, mean life = . dt λ   tdN  t 0  =  dN   t 0     t t0   t0 dN dt dt dN dt dt   = t 0   (N 0 )e t t dt (N 0 )et dt = 1 .   Mean life = 1 λ t 0 k1 k2  B   C (Product), one may show that the In the case of first order consecutive reactions A  concentration of B is given by CB = k1CA0 e –k1t – e –k 2 t  .  (k 2 – k1 )  The usual assumptions are CA = CA0 at t = 0 while CB = CC = 0 at t = 0. Show that CB has its maximum value at  2.303   k2 time t =  k – k  log  k 1   2  1  .  CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Sol. k1C°A dCB  k e –k 2 t – k1e –k1t  =  (k 2 – k1 )  2 dt k 2 e –k 2 t = k 1 e –k1 t equating this zero  k2  k2  k = e(k 2 –k1 ) t n  k  = (k2 – k1) t 1  1   2.303   k2  t =  k – k  log  k  1   2  1 MECHANISM OF A REACTION Reactions can be divided into   Elementary / simple / single step Complex / multi-step ELEMENTARYREACTION  These reaction take place in single step without formation of any intermediate T.S. Ep P.E. Er Reaction coordinates  For elementary reaction we can define molecularity of the reaction which is equal to no of molecules which make transition state or activated complex because of collisions in proper orientation and with sufficient energy  molecularity will always be a natural no 1 = unimolecular one molecule gets excited (like radioactivity) 2.= bimolecular 3 = trimolecular  Molecularly  3 because the probability of simultaneous collision between 4 or more molecules in proper orientation is very low  For elementary reaction there is only single step and hence it is going to be rate determining step so order of an elementary reaction is its molecularity Order of elementary reaction w.r.t. reactant = stoichiometric co-efficient of the reactant H2 + 2 2H (Simple reaction) rate = k [H2] [2] 2H2 + 22 4H ( not elementary ) reaction obtained by multiplying an elementary reaction with some no will not be of elementary nature H2 + Cl2 order = 0 2HCl COMPLEX REACTION :  T.S. Reaction which proceed in more than two steps. or having some mechanism. ( sequence of elementary reaction in which any complex reaction proceeds) E intermediate Reaction coordinates  For complex reaction each step of mechanism will be having its own molecularity but molecularity of net complex reaction will not be defined.  Order of complex reaction can be zero fractions whole no, even negative w.r.t. some species.  Order of reaction or rate law of reaction is calculated with the help of mechanism of the reaction generally using Rate determine step (R.D.S) if given.  Rate law of a reaction is always written in terms of conc. of reactant, products or catalysts but never in terms of conc. of intermediates. The mechanism of any complex reaction is always written in terms of elementary steps, so molecularity of each of these steps will be defined but net molecularity of complex reaction has no meaning. The mechanism of most of the reaction will be calculated or predicted by using mainly the following approximations. CALCULATION OF RATE LAW/ ORDER (A) MECHANISM IN WHICH R.D.S. GIVEN (i) If R.D.S. involves only reactant, product or catalyst on reactant side rate law of R.D.S. = rate law of reaction Ex. Calculate order and rate law of reaction 2NO2 + F2 2NO2F k1  NO2 + F2  NO2 + F NO2F According to RDS with help of mechanism NO2F + F (slow ) (fast ) k2 molecularity = 2 for both Rate = k1 [NO2] [F2] Ex. Calculate rate law 3ClO–  ClO3– + 2Cl– 2ClO– k1 ClO2– + Cl– ClO2– + ClO– k2 ClO3– + Cl– (slow) (fast ) Rate = k1[ClO– ]2 (ii) RDS is having intermediate on reactant side To calculate order, we have to specify [ intermediate] in expression of rate law in terms of conc. of [R], [P] or catalyst with the help of same equilibrium step given in mechanism. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex 2O3  3O2 k1 O3 O2 + O k2 (fast eq. step) (intermediate) O + O3 k 2O2 (slow) From R.D.S., rate = k3 [O3] [ O] According to equilibrium step. 3 k1 [O2 ][O] Keq. = k = [O ] 2 3  k1 [O3 ]  [O] =  k [O ]   2 2   k1 [O3 ]  Rate = k3 [O3] [O] = k3 [O3]  k [O ]  .  2 2  k1 k 3 [O3 ]2 Rate = k [O 2 ] 2 Ex H+ + HNO2 + C6H5NH2 Br C6H5N2+ + 2H2O intermediate  k1 H+ + HNO2 H2 NO2+ k2 (fast) equilibrium step intermediate H2NO + Br + 2 – k3 NOBr + H2O k4 NOBr + C6H5NH2 – Sol. r = k3 [Br–] [H2NO2+] k1 [H 2 NO 2 + ] Keq = k = [H + ][HNO 2 ] 2  k1  [H2NO2+] =  k  [H+] [HNO2]  2 r Ex = (slow) C6H5N + Br + H2O + 2 k1 k 3 [H+] [HNO2] [Br–]. k2 2O3  3O2 O3 O2 + O O2 + O k2 O3 O3 + O k3 2O2 – (fast) Sol. rate = 1 d[O 2 ] 1 d[O3 ] = 3 dt 2 dt d[O3 ] = – k1 [O3] + k2 [O2][O] – k3 [O3] [O] dt d[O 2 ] = k1 [O3] – k2 [O2] [O] + k3 [O3] [O] dt At steady state d [O] =0 dt d[O] = k1 [O3] – k2 [O2] [O] – k3[O3] [O] = 0 dt k1[O3 ] [O] = k [O ] + k [O ] 2 2 3 3 k1 k 3[O3 ]2 k 2 [O2 ]k1[O3 ] d[O3 ] = – k1 [O3] + – k 2 [O 2 ] + k 3[O3 ] k 2 [O 2 ] + k 3[O3 ] dt = – k1 [O3] + k1k 2 [O 2 ][O 3 ] – k1k 3 [O3 ]2 k 2 [O 2 ] + k 3[O3 ] k1k 2 [O 2 ][O 3 ] – k1k 3[O3 ]2 + k1k 2 [O 2 ][O3 ] – k1k 3[O3 ]2 = k 2 [O 2 ] + k 3[O 3 ] = –2k1 k 3[O3 ]2 k 2 [O 2 ] + k 3[O3 ] k1 k 3 [O3 ]2  –1 d   2 dt [O3 ] = k [O ] + k [O ]   2 2 3 3 Rate = – 1 d [O ] 2 dt 3 So, Rate (r) = k1 k 3[O3 ]2 k 2 [O 2 ] + k 3[O3 ] if 3rd step is RDS then k1>> k3 k2 >> k3 r= k1 k 3[O3 ]2 k 2 [O 2 ] + k 3[O3 ] CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Ex H+ + HNO2 + C6H5NH2 k1 H+ + HNO2 H2NO2+ C6H5N2+ + 2H2O H2NO2+ H+ + HNO2 k2 k3 HNO2+ + Br– NOBr + C6H5NH2 Sol. Br NOBr + H2O. k4 C6H5N2+ + H2 O + Br– d[H 2 NO 2 + ] = k1 [H+][HNO2] – k2 [H2NO2+] – k3 [H2 NO2+] [Br–] = 0 dt [H + ]]HNO 2 ] k 2 + k 3 [Br – ] [H2NO2+] = k1 rate = d d [C6 H 5 NH 2 ] = [HNO2 ] dt dt d [C6 H 5 NH 2 ] = k4 [ONBr ] [C6H5NH2] dt d [NOBr ] = – k4 [C6H5NH2][NOBr ] + k3 [Br–] [H2NO2+] = 0 dt k 3[Br – ][H 2 NO 2 + ] [NOBr] = k 4 [C6 H 5 NH 2 ] r = k 4 k 3[Br – ][H 2 NO 2 + ] [C 6H5NH2 ] k 4 [C6 H 5 NH 2 ] r= k1 k 3 [Br – ][H + ][HNO 2 ] k 2 + k 3[Br – ] HYPOTHESIS OF STEADY STATE There are many reactions which involve multi steps and the intermediates (one or more) do not appear in the overall equation, e.g. 2NO + O2  2NO2 Steps involved in the above reaction are NO + O2  NO3 NO + NO3  2NO2 2NO + O2  2NO2 In the above steps, NO3 is an intermediate species and do not appear in the overall balanced equation. Usually these intermediates are very reactive and do not accumulate to any significant extent during the reaction. For hypothetical reaction, A  B The reaction can proceeds in the following steps : A I k1 k2 I B The concentration of the intermediate [I] is much less than the reactant [A] as well as the product [B]. Accordingly the formation of intermediate will start at zero, rises to maximum, and then fall back to zero. If the concentration of intermediate remains small during the reaction, then the curves of reactants, intermediate and product versus t will be given as below : From the plot, it is clear that the slope of the curve for intermediate much less than those for reactants A and products B. It is, therefore d[I] good approximation to take = 0, for each reaction intermediate. dt This is steady state (stationary state) approximation. B A Conc. t RADIOACTIVITY All radioactive decay follows first order kinetics and this is where the similarity ends. This will be explained later in the chapter. We have measured the rate of reaction in chemical kinetics based on the rate of change of concentration of reactants or products. But this procedure will not work for calculating the rate of a radioactive reaction. This is because most of the time the radioactive substance is a solid. Therefore, its concentration would be a constant with time (assuming it to be pure and that the product does not remain with the reactants). Therefore, the rate of radioactive reactions is measured by calculating the rate of change of number of nuclei of the radioactive substance. For a radioactive decay A  B, the rate of reaction is calculated as –dN A = λN A dt where  = decay constant of reaction. NA = number of nuclei of the radioactive substance at the time when rate is calculated. As you can see, the above rate law is very much similar to the rate law of a first order chemical reaction, but all other similarities ceases here. For example unlike a chemical reaction the decay constant () does not depend on temperature. Arrhenius equation is not valid for radioactive decay. –dN A = λN A dt Integrating the differential rate law, we get – Nt t dN A = λ  dt NA N0 0  log N0 = t Nt where N0 = number of nuclei of A, at t = 0 Nt = number of nuclei of A, at t = t  = decay constant The expression can be rearranged to give Nt = N0e–t ..............(1) This suggest that the number of nuclei of radioactive substance A at any instant of time can be calculated, if we know the number of nuclei at t = 0, its decay constant and the time. CHEMICAL KINETICS AND NUCLEAR CHEMISTRY Half-Life Just like a first order reaction, the half-life of radioactive decay is given by t1/2 = 0.693 λ [Note : Let us start with 10 nuclei. If the half-life is 5 minutes, then at the end of first 5 minutes, number of nuclei would be 5. Now, what would be the number of nuclei after next 5 minutes ? Will it be 2.5 or 2 or 3 ? We can clearly see that it cannot be 2.5 and if it is 2 or 3 then it cannot be called as half-life. This dilemma can be overcome by understanding that all formula relating to kinetics are only valid when the sample size is very large and in such a large sample size, a small difference of 0.5 will be insignificant. The fact that radioactive decay follows the exponential law implies that this phenomenon is statistical in nature. Every nucleus in a sample of a radionuclide has a certain probability of decaying, but there is no way to know in advance which nuclei will actually decay in a particular time span. If the sample is large enough, i.e. if many nuclei are present - the actual fraction of it that decays in a certain time span will be very close to the probability for any individual nucleus to decay. To say that, a certain radioisotope has a half-life of 5 hr. signifies that every nucleus of this isotope has a 50 percent chance of decaying in every 5 hr. period. This does not mean probability of 100 percent decaying is 10 hr. A nucleus does not have a memory, and its decay probability per unit time is constant until it actually does decay. A half-life of 5 hr. implies a 75 probability of decay in 10 hr., which increases to 87.5% in 15 hr., to 93.75% in 20 hr., and so on, because in every 5 hr. The probability of decay is 50 percent. Average–Life Time Average life time is defined as the life time of a single isolated nucleus. Let us imagine, a single nucleus which decays in 1 second. Assuming 1 second time interval to be very small, the rate of change of nuclei would be 1/ 1 (because –dN = 1 and dt = 1). We can also see that since –dN –dN = λN , for a single isolated nucleus N = 1, = λ. dt dt Therefore, in this present case,  = 1. Now, let us assume, the same nucleus decays in 2 seconds, we can see that –dN , i.e.  is equal to ½ . You will dt also notice that in the 1st case the nucleus survived for 1 second and in the second case it survived for 2 seconds. Therefore, the life time of a single isolated nucleus is  tav = 1 . λ 1 λ Activity Activity is the rate of decay of a radioactive element. It is represented as 'A' and is equal to N. By no means should activity be confused with rate of change of radioactive nuclei represented by –dN –dN . This is because talks dt dt about the overall change in the number of nuclei in a given instant of time while activity only talks about that change which is decay. For example, if you go to a market with Rs. 50 in your pocket and you spend Rs. 20 in 5 minutes then your rate of change of money in the wallet is Rs. 4/min and in fact the rate of spending the money is also Rs. 4/min. Here, you can see both are same. But if while spending Rs. 20 in 5 minutes, somebody keeps Rs. 10 in your wallet, then the rate of change of money in your wallet would become Rs. 2.5 /min while the rate of spending the money is Rs. 4/min. This implies that as long as the radioactive substance is only decaying the rate of change of nuclei and activity are same and Eq.(1) in terms activity of radioactive substance can be written as At = A0e–t. But if the radioactive substance is also being produced, then dN = rate of production – activity (of dt course it's a different matter that rate of production may or may not be a constant). Specific Activity It is defined as per unit mass of the sample. Let, radioactive sample weighing w g have a decay constant . The w × N 0 , where M = molecular weight of the radioactive substance and N0 number of nuclei in the w g would be M = Avogadro's number. w    λ × × N0  λ × N M  = 0  Specific activity = M w It should be remembered that if a radioactive sample is pure and the product does not remain with reactant, then specific activity is a constant. Units of Activity The unit of radioactivity of a substance is measured as the rate at which it changes into daughter nucleus. It has been derived on the scale of disintegration of radium. Let us consider, 1g of radium (atomic mass = 226 and t1/2 = 1600 yrs) undergoes decay, then Rate of decay of radium =  × Number of nuclei of Ra in 1g = 0.693 1× 6.023×1023 × = 3.7 × 1010 dps = 3.7 × 1010 becquerel 1600 × 365 × 24 × 60 × 60 226 = 1 curie ( 1 Ci = 3.7 × 1010 dps) = 3.7 × 104 Rutherford ( 1Rd = 106 dps) The SI unit of activity is dps or Becquerel. Ex. 210 84 Po decays with emission of -particle to 206 Pb with a half-life period of 138.4 days. If 1g of 210 Po is placed in a 82 84 sealed tube, how much helium will be accumulated in 69.2 days ? Express the answer in cm3 at STP. Sol. 210 84 Po  206 Po + 42He 82 Amount of 210 84 n Po left after 69.2 days can be calculated by applying N = N0 (1/2) n = t/t1/2 = 69.2 1 = 138.2 2 1/ 2 1 N = 1 ×   = 0.702 g 2 Amount of polonium disintegrated = 1 – 0.7072 = 0.2928 g  Moles of polonium in 0.2928 g = 0.2928 210 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 0.2928 210 0.2928 Volume of helium collected = × 22400 = 31.23 cm3 210 Moles of helium atoms formed =  CARBON DATING The cosmic ray generates neutrons in the atmosphere which bombards the nucleus of atmospheric nitrogen to form radioactive 14C hence 14C in the atmosphere has been remaining constant over thousands of years. In living materials, the ratio of 14C to 12C remains relatively constant. When the tissue in an animal or plant dies, assimilation of radioactive 14 C ceased to continue. Therefore, in the dead tissue the ratio of 14C to 12C would decrease depending on the age of the tissue. 14 7 14 6 N + 10n  146 C + 11p C + 14 N + –10e 7 A sample of dead tissue is burnt to give carbon dioxide and the carbon dioxide is analysed for the ratio of 14C to 12 C. From this data, age of dead tissue (plant or animal) can be determined. Age (t) = 2.303 N log 0 λ N  Age = 2.303 × t1/2 (14 C) N log  0  0.693  N  N0 = ratio of 14C/12C in living plant N = ratio of 14C/12C in the wood 2.303× t1/2  A0  log    A  0.693 A0 = Original activity A = Final activity Age = n t 1 Also, N =   N 0  where n = t1/2 2 ROCK DATING It is based on the kinetics of radioactive decay. It is assumed that no lead was originally present in the sample and whole of it came from uranium. Initial no. of mole (N0) = [U] + [Pb] Final no. of mole (N) = [U] [Pb] N 0 [U] + [Pb] = =1+ [U] N [U] t= 2.303 N  log  0  λ N t =  [Pb]  2.303 log 1 + [U]  λ   [Pb]  Also, 1 + = (2)n [U]    t  n =  t1/2   Ex. A sample of uranium mineral was found to contain 206Pb and 238U in the ratio of 0.008 : 1. Estimate the age of the mineral (half-life of 238U is 4.51 × 109 years). Sol. t=  2.303 log 1 + λ  t= 2.303  × t1/2 × log 1 + 0.693  206 Pb   U 238 206 Pb   U 238 Ratio by mass 206Pb : 238U = 0.008 : 1 0.008 1 : Ratio by moles 206Pb : 238U = = 0.0092 206 238 2.303 × 4.51×109 t= log (1 + 0.0092) 0.693 0.0412 2.303 × 4.51×109 ×109 = 5.945 × 107 years = × 0.00397 = 0.693 0.693  If number of neutrons is plotted against the number of protons, the stable nuclei lie within well-defined region called zone of stability. All the nuclei falling outside this zone are invariably radioactive and unstable in nature. Nuclei that fall above the stability zone has an excess of neutrons while those lying below have more protons. These nuclei attain stability by making adjustment in n/p ratio. Number of neutrons (n) STABILITY OF NUCLEI WITH RESPECT TO NEUTRON - PROTON RATIO : 140 120 100 80 60 40 20 0 Unstable region Zone of stability Unstable region n/p=1 20 40 60 80 100 120 140 When (n/p) ratio is higher than that required for stability : Number of protons (p) Such nuclei have tendency to emit -rays (transforming a neutron into proton). n p n : p 1 0 14 6 n  11 p + –10 e (-particle) U  147 N + –10 e 1.33 87 36 1 Kr  87 Rb + –10 e 37 51 36 50 37 When (n/p) ratio is lower than that required for stability : Such nuclei have tendency to increase n/p ratio by adopting any of the following three ways.  By emission of an -particle (natural radioactivity). n   P  298 92 U 234 Th + 24He (-particle) 90  146     92  =1.50  144     90  =1.60 By emission of positron n p 13 7 N  136C + +10 e 6 7 7 6 CHEMICAL KINETICS AND NUCLEAR CHEMISTRY  By K-electron capture 194 79 0 Au + e–1  194 Pt 78 115 116 n   P 79 78 -emission is usually observed in natural radioactive isotopes while emission of positron or K-electron capture is observed in artificial radioactive isotopes. The unstable nuclei continue to emit  or -particle until stable nucleus comes into existence. NUCLEAR FISSION It is a nuclear reaction in which heavy nucleus splits into lighter nuclei of comparable masses with release of large amount of energy by bombardment with suitable sub-atomic particles, i.e. 1 0 n + 235 U 92 56 Ba140 + 36Kr93 + 30n1 54 Xe144 + 38Sr90 + 20n1 55 Cs144 + 37Rb90 + 20n1 If the neutrons from each nuclear fission are absorbed by other 92U235 nuclei, these nuclei split and release even more neutrons. Thus, a chain reaction can occur. A nuclear chain reaction is a self sustaining series of nuclear fissions caused by the previous neutrons released from the previous nuclear reactions. n1 235 92 n 235 92 0 56 n + 23592U 1 0 Ba140 + 10n U 235 92 0 92 Kr + 0 n 93 36 1 0 n 1 U235 +10n 1 92 U U U 235 92 U235 There should be critical amount of the fissionable material to maintain fission chain. This in turn requires, minimum critical mass of the fissionable material. It is the small mass of the fissionable material in which chain reaction can be sustained. If mass is larger than critical mass (supercritical mass), then the number of nuclei that split, multiplies rapidly. An atomic bomb is detonated with small amount of chemical explosive that push together two or more masses of fissionable material to get a supercritical mass. A nuclear fission reactor is a device that permits a controlled chain nuclear fissions. Control rods made of elements such as boron and cadmium, absorb additional neutrons and can therefore, slow the chain reactions. NUCLEAR FUSION It is a nuclear reaction in which two lighter nuclei are fused together to form a heavier nuclei. To achieve this, colliding nuclei must posses enough kinetic energy to overcome the initial force of repulsion between the positively charged core. At very high temperature of the order of 106 to 107 K, the nuclei may have the sufficient energy to overcome the repulsive forces and fuse. Such reactions are therefore also known as thermonuclear reactions. 2 1 H + 13 H   24 He + 10 n + 17.8MeV 2 1 H + 12 H   42 He + 24.9 MeV 1 1 H + 13H   42 He + 20.0 MeV 7 3 Li + 11H   42 He + 17.7 MeV The energy of fusion process is due to mass defect (converted into binding energy). The high temperature required to initiate such reaction may be attained initially through fission process. Hydrogen bomb is based on the principle of fusion reactions. Energy released is so enormous that it is about 1000 times that of atomic bomb. In hydrogen bomb, a mixture of deuterium oxide (D2O) and tritium oxide (T2O) is enclosed in space surrounding an ordinary atomic bomb. The temperature produced by the explosion of the atomic bomb initiates the fusion reaction between 12 H and 13 H releasing huge amount of energy.. It is believed that the high temperature of stars including the sun is due to fusion reactions. E. Salt Peter in 1953, proposed a proton-proton chain reaction. 1 1 H + 11H   12 H + 0+1e + γ 2 1 H + 11H   32 He + γ 3 2 0 He + 11H   42 He + +1 e+γ 411 H   42 He + 20+1 e + 24.7MeV CHEMICAL KINETICS AND NUCLEAR CHEMISTRY 1. Expression for rate constants for reaction of different orders Integrated rate equation Type of reaction Zero order reaction – d[A] = k 0[A] 0 dt Differentiation form First order reaction k1 Second order reaction Unit of rate constant dx k dt 2.303 a log10 t (a – x) k2 = time t life period a 2k 0 -- t1 = Concentration/time–1 2 t1 = –1 2 2.303 b(a – x) log t(a – b) a(b – x) t1 = –1 Differential form Mole litre time –1 3/4 Half-life period 2 0.69 3 K1 t3 =2× 4 1 k2a 0 .6 93 1.382 = k1 k1 t3 = 4 3 k 2a dx = k(a – x) 2 dt Third order reaction k3 = x × (2 a – x) t2a 2 (a – x) 2 2 Differential form –2 Litre mole time –1 t1 = 2 3 2k 3a 2 -- dx = k(a – x)3 dt 2. Some typical linear plots for reactions of different orders : (i) r Zero order r (a0) (ii) (a–x) (iii) Second order (a0)2 1 (a–x) t First order (a0) r Third order (a0)3 Second order t 1 (a–x)2 t1/2 t1/2 t1/2 Zero order (a0) First order log (a–x) Zero order t t1/2 r First order (a0)1 Second order 1 (a0) Third order 1 a 02 Third order t n 3. 1 Amount left after n half-lives =   [A]0 2 No. of half-lives = Total time t 12 st 4. Exponential form of expression for rate constant for reaction of 1 order : [A] = [A]0 e 5. t 12  6. Arrhenius equation for effect of temperature on rate constant, –kt or Ct = C0e –kt 1 [A]n0 –1 –E /RT a k = Ae on log k = log R – Also, Ea 2.303RT dlnk E a = 2 dT RT If k1 and k2 are rate constants at temperature T 1 and T2, then log 7. Ea k2  T1 – T2  = k1 2.303R  T1T  st Examples of reactions of 1 order and their formula for rate constants (i) 2N2O5  4NO2 + O2 ; k = V 2.303 log V – V t  t where V = volume of O2 gas collected at infinite time Vt = volume of O2 gas collected at time t (ii) NH4NO2  2H2O + N2 ; k = V 2.303 log V t  – Vt where V and Vt are volumes of N2 gas collected after infinity time and after time t respectively. (iii) H2O2  H2O + 1 2 O2 ; k = 2.303 VD log t Vt where VD and Vt are the volumes of KMnO4 solution used for titrating a definite volume of the reaction mixture at t = D and at time t respectively. (iv) CH3COOC2H5 + H2O  CH3COOH + C2H5OH k= 2.303 V – VD log t V – Vt CHEMICAL KINETICS AND NUCLEAR CHEMISTRY where VD, Vt and V are the volume of NaOH solution used for titration mixture at zero time, after time t and after infinity respectively. H (v) C12H22O11 + H2O + C6H12O6 + Glucose k= C6H12O6 Fructose 2.303 (r – r ) log 0  t (rt – r ) where r0, rt and r are the polarimetric reading at zero time, after time t and after infinity respectively. 8. Rate law equation for reactions involving parallel reaction k1 B (90%) A k2 Rate = – 9. C (10%) d[A] = k1 [A] + k2 [A] = [k1 + k2] [A] [dt] –kt Degree of dissociation at any time t = (1 – e )