19ElementaryMn2or(1.36)~2niTherefore,atradioactiveequilibrium,theamountofdifferentradioelementspresent will be inversely proportional to their decay constant, or directlyproportional to their half-or average-lifeperiods.Thetimetakenforthecompleteequilibriumtobe established depends onthelifeperiods of elementsinvolved.Thelongerthe averagelifeof anyelement,more slowlyan equilibriumwith itsproductisreached.Problem 1.12The half-life of a radioactive substance 9'x is 20 years.Calculatethe decay constant.In how many years three quarters of the given amount ofsubstancewilldisappear?Solution1year=365×24×60×60=3.15×107sec0.693tu2 = (decay constant)0.693入=20×3.15×107Sincedecayfollowsfirst-orderkinetics,tunis independent ofa i.e.initialconcentrations.50%ofsubstancedecaysin20years,next50%ofremaining,i.e.25% will decay in 20 years.Therefore, 75% will decay in 40 years.Problem1.13Aradioactiveelementgives2000countsperminatagiventime.Afteronehour,countswerefoundtobe750permin.Whatishalf-lifeof theelement?SolutionThedecay constantNo0.6930.303入=logN112Noα2000andN×750t =1 hour =60 min0.303 10g20000.69375060tu2orf1/2=42.5minProblem1.14Half-lifeof radioactive elementis 2800years.Howmanyatomsoftheelementarerequiredtoproduceanaverageof 10beta emissionsper hour?
Elementary 19 or λ λ 1 2 2 1 = . . . n n (1.36) Therefore, at radioactive equilibrium, the amount of different radio elements present will be inversely proportional to their decay constant, or directly proportional to their half- or average-life periods. The time taken for the complete equilibrium to be established depends on the life periods of elements involved. The longer the average life of any element, more slowly an equilibrium with its product is reached. Problem 1.12 The half-life of a radioactive substance 91X is 20 years. Calculate the decay constant. In how many years three quarters of the given amount of substance will disappear? Solution 1 year = 365 × 24 × 60 × 60 = 3.15 × 107 sec t1/2 = 0.693 λ (decay constant) λ = 0.693 20 3.15 107 × × Since decay follows first-order kinetics, t1/2 is independent of a i.e. initial concentrations. 50% of substance decays in 20 years, next 50% of remaining, i.e. 25% will decay in 20 years. Therefore, 75% will decay in 40 years. Problem 1.13 A radioactive element gives 2000 counts per min at a given time. After one hour, counts were found to be 750 per min. What is half-life of the element? Solution The decay constant λ = 0.303 log = 0 0.693 1/2 t N N t ⎛ ⎝ ⎞ ⎠ N0 ∝ 2000 and N ∝ 750 t = 1 hour = 60 min 0.303 60 log 2000 750 = 0.693 1/2 t or t1/2 = 42.5 min Problem 1.14 Half-life of radioactive element is 2800 years. How many atoms of the element are required to produce an average of 10 beta emissions per hour?
20Chemical KineticsandReactionDynamicsSolutionfin=2800×365×24=2.45×107 hours0.693= 0.28 × 10-7 hour-11=2.45×107-dt-dN/dtorn=and1-dNldt=10(given)10(atoms/hour)0.28 ×10-7 hour- = 35.7 × 107 = 3.57 ×10°atomsn=1.9 Second OrderReactionsLetustakeasecondorderreactionA+Bk>Productsin which initial concentration of each reactant A and B is same, say,a moldm-3. If after time t, x moles of each reactant is reacted, the concentration ofeachwillbe(a-x)andrateof reaction wouldbe=k (a-x)(a-x)= k (a-x)dtRearrangingtheequations,wehavedx=kdt(a-x)2Onintegrating,itgives1=kt +Z(a - x)The integration constant Z can be obtained by putting x =Owhen t=OZ=laOnsubstitutingthevalueof Z,therateequationbecomes1=kt+ 1(1.37)(a-x)a1Yk=or(1.38)a(a-x)
20 Chemical Kinetics and Reaction Dynamics Solution t1/2 = 2800 × 365 × 24 = 2.45 × 107 hours λ = 0.693 2.45 10 = 0.28 10 hour 7 –7 –1 × × – = dN dt λn or n dN dt = – / λ and – dN/dt = 10 (given) n = 10 (atoms/hour) 0.28 10 hours = 35.7 10 = 3.57 10 atoms –7 –1 7 8 × × × 1.9 Second Order Reactions Let us take a second order reaction A + B →k Products in which initial concentration of each reactant A and B is same, say, a mol dm–3. If after time t, x moles of each reactant is reacted, the concentration of each will be (a – x) and rate of reaction would be dx dt = k (a – x)(a – x) = k (a – x) 2 Rearranging the equations, we have dx a x kdt ( – ) = 2 On integrating, it gives 1 ( – ) = + a x kt Z The integration constant Z can be obtained by putting x = 0 when t = 0 Z a = 1 On substituting the value of Z, the rate equation becomes 1 ( – ) = + 1 a x kt a (1.37) or k t x aa x = 1 = ( – ) (1.38)
Elementary21: The unit of rate constant is (conc.)-' (time)-l or mol- dm’ sec-!. The half-life period ti2 is obtained by putting t = tu2 and x = a/2 inequation (1.38)asa/21k=a(a-a/2)t121(1.39)ortun=kaThus thehalf-lifeperiod ofa second orderreaction is inverselyproportionalto the initial concentration of the reactant.:Accordingtoequation(1.37),itisobserved that aplotof 1/(a-x)(xr-versus time should be linear withSlope = kslope k and a positive intercept31/a1/a (Fig. 1.4).TimeThus,in case of a second-orderreaction,aplotof 1/(conc.ofreactant)Fig. 1.4Second order plot.versus time would always be linearwith a positive intercept.Second Order Reaction with Reactants having Different InitialConcentrationsLet thereactionABxkProductsb0InitiallyaxAttime t,(a -x)(b - x)The rate of reaction is represented asdx=k(a-x)(b-x)dtdx= kdt(1.40)or(a-x)(b-x)Nowwe use the method of partial fractions to writethe left side as sumoftwosimpleterms.Let1pq(a-x)(b-x)(a-x)(b -x)where p and q are evaluated by using a common denominator and equatingcoefficients of likepowers of x in numerator.Thevalues of p and qmaybedeterminedasp=1/(b-a)and q=-1/(b-a).Equation(1.40)thenbecomesdxdx(1.41)=kdt(b-a)(a-x)(b-a)(b-x)
Elementary 21 • The unit of rate constant is (conc.)–1 (time)–1 or mol–1 dm3 sec–1 • The half-life period t1/2 is obtained by putting t = t1/2 and x = a/2 in equation (1.38) as k t a aa a = 1 /2 ( – /2) 1/2 ⋅ or t ka 1/2 = 1 (1.39) Thus the half-life period of a second order reaction is inversely proportional to the initial concentration of the reactant. Fig. 1.4 Second order plot. with a positive intercept. Second Order Reaction with Reactants having Different Initial Concentrations Let the reaction A + B ⎯ → ⎯k Products Initially a b 0 At time t, (a – x) (b – x) x The rate of reaction is represented as dx dt = ( – )( – ) ka x b x or dx a xb x kdt ( – )( – ) = (1.40) Now we use the method of partial fractions to write the left side as sum of two simple terms. Let 1 ( – )( – ) = ( – ) + a xb x ( – ) p a x q b x where p and q are evaluated by using a common denominator and equating coefficients of like powers of x in numerator. The values of p and q may be determined as p = 1/(b – a) and q = –1/(b – a). Equation (1.40) then becomes dx b aa x dx b ab x kdt ( – )( – ) – ( – )( – ) = (1.41) 1 ( – ) a x 1/a Slope = k Time • According to equation (1.37), it is observed that a plot of 1/(a – x) versus time should be linear with slope k and a positive intercept 1/a (Fig. 1.4). Thus, in case of a second-order reaction, a plot of 1/(conc. of reactant) versus time would always be linear
22Chemical KineticsandReactionDynamicsNowthe integration of eachterm is simpleand the result is1/(b-a) ln al(a-x)-1/(b-a) lnbl(b-a)Inbl(b-x)=ktwhich reduces to11[(b-x)a][b(a-x)k=TInorI1(b-a)rb(a-x)(a-b)t[a(b-x)[(b-x)al2.303ork=log(1.42)(b-a)r[b(a-x)Thus,equation(1.42)isused to obtaintherate constantfora second-orderreactioninwhichboththereactantshavedifferentinitialconcentration.Whether the experimental data satisfy this expression can be checked byplotting log (a(b-x)/b(a-x)) against time,which should be a straight linepassing throughthe origin.The slope of straight line gives thevalue of2.303/k(b-a)fromwhich thevalueof rateconstantk canbeevaluatedThehalf-lifeperiod in this casecanonlybedetermined if thereactants aretakenin stoichiometricamounts.Let a and bbethe initial concentrations of A and B,respectively,for thereactionIA+mB→ProductsmA+B→ProductsTb0InitiallyamAttimet a-x1The rate,d=k(a-xdIfa and b are in stoichiometric amounts,i.e.b= (m/l)a,the above equationreduces to禁= k(ml)(a - x)2dtRearranging and integrating,the rate constant and half-life period may beobtainedas111k:(1.43)t(m/l)((a-x)aand1/2 = /mka(1.44)
22 Chemical Kinetics and Reaction Dynamics Now the integration of each term is simple and the result is 1/(b – a) ln a/(a – x) – 1/(b – a) ln b/(b – a) ln b/(b – x) = kt which reduces to k b at b xa ba x = 1 ( – ) ln ( – ) ( – ) ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ or k a bt ba x ab x = 1 ( – ) ln ( – ) ( – ) ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ or k b at b xa ba x = 2.303 ( – ) log ( – ) ( – ) ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ (1.42) Thus, equation (1.42) is used to obtain the rate constant for a secondorder reaction in which both the reactants have different initial concentration. Whether the experimental data satisfy this expression can be checked by plotting log {a(b – x)/b(a – x)} against time, which should be a straight line passing through the origin. The slope of straight line gives the value of 2.303/k(b – a) from which the value of rate constant k can be evaluated. The half-life period in this case can only be determined if the reactants are taken in stoichiometric amounts. Let a and b be the initial concentrations of A and B, respectively, for the reaction lA + m B → Products A + m l B → Products Initially a b 0 At time t a – x m l x ⎛ ⎝ ⎞ ⎠ x The rate, dx dt ka x b m l = ( – ) – x ⎛ ⎝ ⎞ ⎠ If a and b are in stoichiometric amounts, i.e. b = (m/l)a, the above equation reduces to dx dt = ( / )( – ) k ml a x 2 Rearranging and integrating, the rate constant and half-life period may be obtained as k t ml a x a = 1 ( /) 1 ( – ) – ⎛ 1 ⎝ ⎞ ⎠ (1.43) and t1/2 = l/mka (1.44)
Elementary23Problem1.15Hydrolysis ofester, inpresence ofNaOH, wasfollowedkeepingtheirconcentration same at 0.02mol dm-3.The specific rateconstant at35°Cwasfoundtobe5.55mol-'dmmin-l.Whatproportionofesterwillbehydrolysed in30min?Alsocalculatethetimefor30%decompositionofester.Solution The rate constant for a second-order reaction when the initialconcentrationsofbothreactants aresameisgivenasa = 0.02 mol dm~31xk=t(a-x)k= 5.55mol-l dm2min-l(i)Letffractiondecomposein30min.Thenx=afaff11k=30(a(a-af)30a(1-f)f1or5.55 = 30 × 0.02(1-f)f=5.55×30×0.02=3.33-f=0.76or76%This is the integrated rateequationfor a second-order reaction.Problem1.16ForthereactionA+B→C+Dthe following data were obtained:0Time (sec)1782755308601500[A] × 10′ (mol dm-3)9.88.98.68.07.36.53.04.84.03.72.31.5[B] ×103 (mol dm-3)Calculatethe rate constant and confirmthat reaction is of second order.Howtherateconstantcanbeobtainedgraphically?Solutionb(a-x)2.303k=1oga(b-x)t(a-b)a = 9.8 × 10-3 mol dm-3b = 4.8 × 10-3 mol dm-3
Elementary 23 Problem 1.15 Hydrolysis of ester, in presence of NaOH, was followed keeping their concentration same at 0.02 mol dm–3. The specific rate constant at 35°C was found to be 5.55 mol–1dm3 min–1. What proportion of ester will be hydrolysed in 30 min? Also calculate the time for 30% decomposition of ester. Solution The rate constant for a second-order reaction when the initial concentrations of both reactants are same is given as k t x a x a k = 1 ( – ); = 0.02 mol dm = 5.55 mol dm min –3 –1 3 –1 (i) Let f fraction decompose in 30 min. Then x = af k af a a af a f f = 1 30 ( – ) = 1 30 (1 – ) ⎛ ⎝ ⎞ ⎠ or 5.55 = 1 30 0.02 × (1 – ) × f f f 1 – f = 5.55 × 30 × 0.02 = 3.33 f = 0.76 or 76% This is the integrated rate equation for a second-order reaction. Problem 1.16 For the reaction A + B → C + D the following data were obtained: Time (sec) 0 178 275 530 860 1500 [A] × 103 (mol dm–3) 9.8 8.9 8.6 8.0 7.3 6.5 [B] × 103 (mol dm–3) 4.8 4.0 3.7 3.0 2.3 1.5 Calculate the rate constant and confirm that reaction is of second order. How the rate constant can be obtained graphically? Solution k ta b ba x ab x = 2.303 ( – ) log ( – ) ( – ) ⎛ ⎝ ⎞ ⎠ a = 9.8 × 10–3 mol dm–3 b = 4.8 × 10–3 mol dm–3