24Chemical KineticsandReactionDynamicskTime(sec)log(a-x)log(b - x)b(a -x)/a(b -x)3.99123.68120-3.94203.60200.844×1041781.0783.93403.56820.939 ×10-2751.1383.90303.47701.007×105301.3063.86333.36171.025×108601.5543.81253.17601.003×102.1221500Average = 0.9636 × 10kalmostremains constant.Therefore,thereaction is of second order.Problem1.17Areactantreacts30%in30min.Ifthereactionfollowsasecondorderkinetics,findrateconstantandremainingconcentrationofreactant after 60min.SolutionForasecondorderreactionk=tia-raLet a =100, x = 30 and a - x =70.Putting the values in above,we get111[1=1.42× 10- mol-dm"sec-1k=30[707000100111k =Further,100]60la-x1[100 - (a -x)1.42×10-60(a-x)1001.42×10-4×60×100(a-x)=100-(a-x)85.20 (a-x)+ (a-x)=10086.20 (a-x)=100100=1.16,i.e.1.16%concentrationof reactantwillremain(a-x)=86.20after 60 min.Reduction of OrderLet us consider a second orderreaction which isfirst orderwith respect toeachreactantandthereactantshavedifferentinitialconcentrations.A + B k>Productsab
24 Chemical Kinetics and Reaction Dynamics Time(sec) log(a – x) log(b – x) b(a – x)/a(b – x) k 0 3 .9912 3 .6812 – – 178 3 .9420 3 .6020 1.078 0.844 × 10–4 275 3 .9340 3 .5682 1.138 0.939 × 10–4 530 3 .9030 3 .4770 1.306 1.007 × 10–4 860 3 .8633 3 .3617 1.554 1.025 × 10–4 1500 3 .8125 3 .1760 2.122 1.003 × 10–4 Average = 0.9636 × 10–4 k almost remains constant. Therefore, the reaction is of second order. Problem 1.17 A reactant reacts 30% in 30 min. If the reaction follows a second order kinetics, find rate constant and remaining concentration of reactant after 60 min. Solution For a second order reaction k ta x a = 1 1 – – 1 [ ] Let a = 100, x = 30 and a – x = 70. Putting the values in above, we get k = 1 30 1 70 – 1 100 = 1 7000 = 1.42 10 mol dm sec –4 –1 3 –1 [ ] × Further, k a x = 1 60 1 – – 1 [ ] 100 1.42 10 = 1 60 100 – ( – ) ( – )100 –4 × ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ a x a x 1.42 × 10–4 × 60 × 100 (a – x) = 100 – (a – x) 85.20 (a – x) + (a – x) = 100 86.20 (a – x) = 100 ( – ) = 100 86.20 a x = 1.16, i.e. 1.16% concentration of reactant will remain after 60 min. Reduction of Order Let us consider a second order reaction which is first order with respect to each reactant and the reactants have different initial concentrations. A + B Products a b →k
25ElementaryTherateconstantforthis reactionis givenby[a(b-x)2.303k=logb(a-x)t(b-a)If oneof the reactants,sayB,has a veryhighconcentration in comparisonto that of A, i.e. b >a then, the following conditions would be valid:b-a=bandb-x=b(becausexdepends onboth Aand Band is small as a is small).Therefore,theequationof rateconstantreducesto1 2.303,k=(1.45)10sb(a-x)This rate expression is similar to the integrated expression for a first orderreaction, only the magnitude of rate constant is changed by (1/b)times.Therefore, the reaction is said to be pseudo-first order. The reaction is firstorder with respect to reactant A, which is present at low concentration.TheorderwithrespecttoBis arrangedtobezerobyvirtueof itshighconcentrationThus, it may be concluded that if any reactant is present at a very highconcentration, it does not affect the rate of reaction.The rate will dependonly on those reactants, which are present at low concentrations.This isbecause,ifreactantBistaken inverylargeconcentrationcomparedtootherreactant A, then even if all of the reactant A is used up in the reaction, therewill be very little decrease in the concentration of B.Therefore, the reactionistakingplaceundertheconditions,wheretheconcentrationofBispracticallyconstant.For this reason,the rate of the reaction is a function of theconcentration of only the reactant, which is present at low concentration.Usually,to maintaintheconcentration of onereactant constantduring thereaction, its concentration should be taken in at least ten-fold excess of theotherreagent.Second Order Reaction When Concentrations of Reactants Differ OnlySlightlyIf the concentration of two reactants differ only slightly, i.e. by a smallamount, (a-x)/(b-x)will almost be unity and,therefore,equationb(a-x)1k=t(a-b)a(b-x)cannot be used.In this case, an equation is obtained by expansion of thelogarithmLet a = d+ S and b = d - S, where d is the mean initial concentration, i.e.d= (a +b)/2and 2s is excess of concentration of a over that of b
Elementary 25 The rate constant for this reaction is given by k tb a ab x ba x = 2.303 ( – ) log ( – ) ( – ) ⎧ ⎨ ⎩ ⎫ ⎬ ⎭ If one of the reactants, say B, has a very high concentration in comparison to that of A, i.e. b a then, the following conditions would be valid: b – a ≈ b and b – x ≈ b (because x depends on both A and B and is small as a is small). Therefore, the equation of rate constant reduces to k b t a a x = 1 2.303 log ( – ) (1.45) This rate expression is similar to the integrated expression for a first order reaction, only the magnitude of rate constant is changed by (1/b) times. Therefore, the reaction is said to be pseudo-first order. The reaction is first order with respect to reactant A, which is present at low concentration. The order with respect to B is arranged to be zero by virtue of its high concentration. Thus, it may be concluded that if any reactant is present at a very high concentration, it does not affect the rate of reaction. The rate will depend only on those reactants, which are present at low concentrations. This is because, if reactant B is taken in very large concentration compared to other reactant A, then even if all of the reactant A is used up in the reaction, there will be very little decrease in the concentration of B. Therefore, the reaction is taking place under the conditions, where the concentration of B is practically constant. For this reason, the rate of the reaction is a function of the concentration of only the reactant, which is present at low concentration. Usually, to maintain the concentration of one reactant constant during the reaction, its concentration should be taken in at least ten-fold excess of the other reagent. Second Order Reaction When Concentrations of Reactants Differ Only Slightly If the concentration of two reactants differ only slightly, i.e. by a small amount, (a – x)/(b – x) will almost be unity and, therefore, equation k ta b ba x ab x = 1 ( – ) ln ( – ) ( – ) cannot be used. In this case, an equation is obtained by expansion of the logarithm. Let a = d + S and b = d – S, where d is the mean initial concentration, i.e. d = (a + b)/2 and 2S is excess of concentration of a over that of b
26Chemical KineticsandReactionDynamicsThis can be written in terms of S and d as2kSt=Ind-x) + In(1 - 司) - In (1 + -In二YOnexpandingthelogarithmterms,theequationmaybeobtainedas53S2S2kSt2(d -x)2 3(d -x)3(d -x)53$3s3(d - x)3(d -x)2(d - x)252S3-景号+S25(1.46)2d23d3adOn collecting the terms and arranging,the equation becomes1LS2kt =(1.47)(d-x)a-X)daWhenSis zero,theequationreducesto1(1.48)kt=(d-x)This is equivalentto equation used when boththereactants have same initialconcentrations,i.e.t(a-x) -Equation (1.48)may be used todeterminethe value of rate constant whenthe concentrationsof tworeactants differ onlyslightly(S<<d).Here,aplotof 1/(d-x)against time will be linearwith a slope equal tok.Second Order Autocatalytic ReactionWhen a product, formed in a second order reaction, acts as a catalyst oreffects the rate of reaction, the reaction is known as autocatalytic reaction.Forexample,theacid catalysed hydrolysis of various estersand similarcompoundsandvariousbiochemicalprocesses.Letus taketheexampleof hydrolysisof an esterinpresenceof HClInitially at time t Ester + H,O HCl Acid + Alcohol00abcx(a-x)x(b-x)Since, water being in excess, its concentration may be taken constant.Again concentration of acid (HCl)is c,theratewill depend on (c+x),i.e
26 Chemical Kinetics and Reaction Dynamics This can be written in terms of S and d as 2 ln 1 + – – ln 1 – – kSt = + ln 1 – – ln 1 + S d x S d x S d S d ( )( )( )( ) On expanding the logarithm terms, the equation may be obtained as 2 ( – ) – 2( – ) – 3( – ) + . . . 2 2 3 3 kSt = S d x S d x S d x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ – – ( – ) – 2( – ) – 3( – ) + . . . 3 2 3 3 S d x S d x S d x ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + – – 2 – 3 + . . . – – 2 – 3 + . . . 2 2 3 3 2 2 3 3 S d S d S d S d S d S d ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (1.46) On collecting the terms and arranging, the equation becomes kt dx d S ax d = 1 ( – ) – 1 + 3 1 ( – ) – 1 + . . . 2 3 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (1.47) When S is zero, the equation reduces to kt dx d = 1 ( – ) – 1 (1.48) This is equivalent to equation used when both the reactants have same initial concentrations, i.e. kt C C = 1 – 1 0 ⎛ ⎝ ⎞ ⎠ or 1 ( – ) – 1 a x a ⎛ ⎝ ⎞ ⎠ Equation (1.48) may be used to determine the value of rate constant when the concentrations of two reactants differ only slightly (S << d). Here, a plot of 1/(d – x) against time will be linear with a slope equal to k. Second Order Autocatalytic Reaction When a product, formed in a second order reaction, acts as a catalyst or effects the rate of reaction, the reaction is known as autocatalytic reaction. For example, the acid catalysed hydrolysis of various esters and similar compounds and various biochemical processes. Let us take the example of hydrolysis of an ester in presence of HCl Initially at time t Ester (–) + H O (–) Acid 0 + Alcohol 0 2 HCl a a x b b x c x x ⎯ →⎯ Since, water being in excess, its concentration may be taken constant. Again concentration of acid (HCl) is c, the rate will depend on (c + x), i.e
Elementary27acid present initially+acid formed at any time t.Thus, the rate of reactionis given by些=k(a - x)(c+x)dtOn rearranging and integrating,wegeta(c +x)2.303k=-1og c(a-x)(1.49)t(a+c)Equation(1.49)canbeusedforcalculatingtherateconstantfora second-order autocatalytic reaction.When the concentration (c+ x) isplotted againsttimet,aS-shapedcurve is obtained. (Fig. 1.5). This[x+ 3] curveischaracteristicofautocatalyticreactions and manygrowthprocesses.It can be seen from equation (1.49)Timethatratewillbemaximum whenFig. 1.5 Plot of (c + x) versus time.(a-x)=(c+x)or x=(a-c)/2Thus,ifcis small incomparisontoa,themaximumratewillbeobtainedwhentheconcentrationahasdecreasedby50%.Again the rate constant atthepoint, where therate is maximum,canbeobtainedbysubstituting c=(a-2x)in equation (1.49).Thus, the ratewillbe given as2.303a(1.50)k=1og (a-2x)2t(a-x)Problem 1.18A second order reaction with initial concentration of eachreactant as 0.5mol dm-3was carried out in presence of acid as catalyst.AtpH4.0thehalf-lifeofreactionwas foundtobe 60min.Calculatetheobservedandtruerateconstantforthereaction.SolutionRate=kurue[H][A][B] =kobs[A][B],when kobs=ktrue [H+][A]o = [B]o = 0.5 mol dm-3[H*] = 10-pPH = 10- mol dm-3For a second orderreaction111/2 =kobs × [initial conc.]
Elementary 27 acid present initially + acid formed at any time t. Thus, the rate of reaction is given by dx dt = ( – )( + ) ka x c x On rearranging and integrating, we get k ta c ac x ca x = 2.303 ( + ) – log ( + ) ( – ) (1.49) Equation (1.49) can be used for calculating the rate constant for a secondorder autocatalytic reaction. When the concentration (c + x) is plotted against time t, a S-shaped curve is obtained. (Fig. 1.5). This curve is characteristic of autocatalytic reactions and many growth processes. It can be seen from equation (1.49) that rate will be maximum when (a – x) = (c + x) or x = (a – c)/2 Thus, if c is small in comparison to Fig. 1.5 Plot of (c + x) versus time. Time [c + x] a, the maximum rate will be obtained when the concentration a has decreased by 50%. Again the rate constant at the point, where the rate is maximum, can be obtained by substituting c = (a – 2x) in equation (1.49). Thus, the rate will be given as k ta x a a x = 2.303 2 ( – ) log ( – 2 ) (1.50) Problem 1.18 A second order reaction with initial concentration of each reactant as 0.5 mol dm–3 was carried out in presence of acid as catalyst. At pH 4.0 the half-life of reaction was found to be 60 min. Calculate the observed and true rate constant for the reaction. Solution Rate = ktrue [H+ ][A][B] = kobs [A][B], when kobs = ktrue [H+ ] [A]0 = [B]0 = 0.5 mol dm–3 [H+ ] = 10–pH = 10–4 mol dm–3 For a second order reaction t k 1/2 obs = 1 [initial conc.] ×
28Chemical Kinetics and Reaction Dynamics160 × 0.5 mol’dm* min = 0.033kobs== 3.33 × 10-2 mol-ldmmin-lkabs --3.33×10-210- mol dm3 mol-'dm'min-kin[H]= 3.33 × 102mol-2dm°min-l1.10ThirdOrderReactionsAthirdorderreactioncanbetheresult of thereactionofa singlereactant,tworeactantsorthreereactants.Ifthetwoorthreereactantsareinvolvedinthereactiontheymayhavesameordifferentinitialconcentrations.Dependinguponthe conditionsthedifferential rateequation maybeformulated andintegratedtogivetherateequation.Insomecases,therateexpressionshavebeen given as follows.(a) Where three reactants are involved with same initial concentrationsConsiderthereactionk> ProductsBcInitiallyattimetA++aOx0(a-x)(a-x)(a-x)Thendx=k(a-x)3Rate =dtOn rearranging and integrating the equation, We get11k=(1.51)2((a-x)2awhichcanbeusedforcalculatingtherateconstantof athird-orderreaction,whenallthereactantshavesameinitial concentrations..Units of rate constant is (conc.)-2 (time)-or generally mol-2 lit'sec-l ormol-? dme sec-l..Half-life period maybe obtained from equation(1.51)by substitutingt= ti/2when x=al2.Thus,3(1.52)11/2=2ka?Therefore, in case of third order reactions the half-life period is inverselyproportionaltothesquareofinitial concentrationEquation(1.51)canberearrangedas
28 Chemical Kinetics and Reaction Dynamics kobs = 1 60 0.5 × mol–1dm3 min–1 = 0.0333 = 3.33 × 10–2 mol–1dm3 min–1 k k true obs + –2 –4 –3 –1 3 –1 = [H ] = 3.33 10 10 mol dm mol dm min × = 3.33 × 102 mol−2 dm6 min−1 1.10 Third Order Reactions A third order reaction can be the result of the reaction of a single reactant, two reactants or three reactants. If the two or three reactants are involved in the reaction they may have same or different initial concentrations. Depending upon the conditions the differential rate equation may be formulated and integrated to give the rate equation. In some cases, the rate expressions have been given as follows. (a) Where three reactants are involved with same initial concentrations Consider the reaction: Initially at time t A (–) + B (–) + C (–) Products a 0 a x a a x a a x x →k Then Rate = = ( – ) dx 3 dt ka x On rearranging and integrating the equation, We get k t ax a = 1 2 1 ( – ) – 1 2 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (1.51) which can be used for calculating the rate constant of a third-order reaction, when all the reactants have same initial concentrations. • Units of rate constant is (conc.)–2 (time)–1 or generally mol–2 lit2 sec–1 or mol–2 dm6 sec–1. • Half-life period may be obtained from equation (1.51) by substituting t = t1/2 when x = a/2. Thus, t ka 1/2 2 = 3 2 (1.52) Therefore, in case of third order reactions the half-life period is inversely proportional to the square of initial concentration. • Equation (1.51) can be rearranged as