4 Chemical KineticsandReactionDynamicsThus, the rate constant is the rate of reaction when concentrations of thereactants areunity.Therate constantundertheseconditions is alsoknown asthe specific rate or the rate coefficient.Therate constant for any reaction canbedetermined(i) eitherbymeasuring the rate of the reaction at unit concentrations ofthe reactants.(i)or byknowing the rate at any concentration of reactant usingtherelationRate constant =Rate/[A]"[B]b(1.10)The rate constant is measured in units of moles dm-3 sec-'(molesdm-3)nwhere n =a+b.Timemay also be in minutes or hours.It should be notedthat in case where the reaction is slow enough, the thermal equilibrium willbemaintained duetoconstantcollisionsbetweenthemolecules andkremainsconstant at a given temperature.However, if the reaction is very fast the tailpart of theMaxwell-Boltzmanndistribution willbedepleted sorapidlythatthermal equilibriumwillnotbere-established.Insuchcasesrateconstantwill not trulybe constant and it should be called a rate coefficient.1.3OrderandMolecularityFor reactionαA +βB+..→Productrateofreaction isproportional toαthpowerofconcentration ofA,totheβthpowerof concentration of Betc.,i.e.Rate = k [A]~ [Bjβ.(1.11)Then the reaction would be said to be αt order with respect to A,βthorderwith respect toB, ...andtheoverall orderof reactionwouldbeα+β+Thus,orderof reactionwithrespecttoa reactant is thepowertowhichthe concentration of the reactant israised intothe rate law,and the overallorder of reaction is the sum of the powers of the concentrations involved inthe rate law.Theterm'molecularity is the sumof stoichiometric coefficientsof reactantsinvolvedinthestoichiometricequationofthereaction.Forexample,areactionwhosestoichiometricequationis2A+3B= 3C+2Dthe stoichiometric coefficientof AandB are2 and 3,respectively,andtherefore, the molecularitywould be2+3=5.There is not necessarily a simple relationship between molecularityandorder of reaction.For differentiating between molecularity and order of areaction, let us consider some examples.For the reaction,A+2B→P,themolecularity is1+2=3.If thereaction
4 Chemical Kinetics and Reaction Dynamics Thus, the rate constant is the rate of reaction when concentrations of the reactants are unity. The rate constant under these conditions is also known as the specific rate or the rate coefficient. The rate constant for any reaction can be determined (i) either by measuring the rate of the reaction at unit concentrations of the reactants. (ii) or by knowing the rate at any concentration of reactant using the relation Rate constant = Rate/[A]a [B]b (1.10) The rate constant is measured in units of moles dm–3 sec–1/(moles dm–3) n , where n = a + b. Time may also be in minutes or hours. It should be noted that in case where the reaction is slow enough, the thermal equilibrium will be maintained due to constant collisions between the molecules and k remains constant at a given temperature. However, if the reaction is very fast the tail part of the Maxwell-Boltzmann distribution will be depleted so rapidly that thermal equilibrium will not be re-established. In such cases rate constant will not truly be constant and it should be called a rate coefficient. 1.3 Order and Molecularity For reaction αA + βB + . . . → Product rate of reaction is proportional to αth power of concentration of A, to the βth power of concentration of B etc., i.e. Rate = k [A]α [B]β . . . (1.11) Then the reaction would be said to be αth order with respect to A, βth order with respect to B, . . . and the overall order of reaction would be α + β + . . . . Thus, order of reaction with respect to a reactant is the power to which the concentration of the reactant is raised into the rate law, and the overall order of reaction is the sum of the powers of the concentrations involved in the rate law. The term ‘molecularity’ is the sum of stoichiometric coefficients of reactants involved in the stoichiometric equation of the reaction. For example, a reaction whose stoichiometric equation is 2A + 3B == 3C + 2D the stoichiometric coefficient of A and B are 2 and 3, respectively, and, therefore, the molecularity would be 2 + 3 = 5. There is not necessarily a simple relationship between molecularity and order of reaction. For differentiating between molecularity and order of a reaction, let us consider some examples. For the reaction, A + 2B → P, the molecularity is 1 + 2 = 3. If the reaction
5Elementaryoccurs in a single step the order of reaction with respect to Awould be oneand order with respect to B would be two,giving overall order of reaction 3.Thus the molecularity and order would be same.However, if the reactionoccurs in two different steps giving overall same reaction, e.g.(a) A+B=I→SlowI+B=P→fast(b)A+2B→PNowtherate of reaction will begoverned byonlyslow step (a)and order ofreaction wouldbeone withrespectto eachreactant,Aand B,givingoverallorder two.And,therefore,the order and molecularitywill be different.The inversion of cane sugar isC12H22O+H,0→CH120+CH1206and therate of inversion is given byRate = k [Sucrose] [H,O](1.12)This reaction seems to be second order, i.e.first order with respect to eachsucrose and H,O.The [HO] is also constant as it is used as solvent andpresent in large amount.Therefore, the reaction is only first order withrespect to sucrose.The hydrolysis of ester in presence of acid is first order reaction (keepingcatalyst constant)CH;COOC,Hs + H2O _H+l, CH;COOH + C,H,OHSince [H,O] remain constant as in case of inversion of cane sugar, it does noteffect the rate of reaction and reaction is simplyfirst order with respecttoester.However,thehydrolysis of ester in presence of alkaliCH,COOC,H, + NaOH→ CH,COONa +C,H,OHis second order being first order withrespect toboth ester and NaOH.Whilethe molecularity of the reaction in each case, i.e. in hydrolysis of ester inpresence of acid as well as in presence of alkali, is two.The reactions, in which molecularity and order are different due to thepresenceofoneofthereactant in excess,areknown aspseudo-orderreactions.Theword (pseudo)is alwaysfollowed byorder.For example,inversion ofcane sugar ispseudo-first orderreaction.The molecularity will always be a whole integer while order may be aninteger,fraction or even a negative number.Molecularity is a theoreticalconcept,whereasorderis empirical.Molecularity is,therefore,lesssignificantas far askinetic studies are concerned.Theorderofreactionprovidesthebasisforclassifyingreactions.Generally,the order of reaction can be anywhere between zero and three.Reactionshaving order three and above are very rare and can be easily counted
Elementary 5 occurs in a single step the order of reaction with respect to A would be one and order with respect to B would be two, giving overall order of reaction 3. Thus the molecularity and order would be same. However, if the reaction occurs in two different steps giving overall same reaction, e.g. (a) A + B = I → Slow (b) I + B = P fast A + 2B P → → Now the rate of reaction will be governed by only slow step (a) and order of reaction would be one with respect to each reactant, A and B, giving overall order two. And, therefore, the order and molecularity will be different. The inversion of cane sugar is C12H22O11 + H2O → C6H12O6 + C6H12O6 and the rate of inversion is given by Rate = k [Sucrose] [H2O] (1.12) This reaction seems to be second order, i.e. first order with respect to each sucrose and H2O. The [H2O] is also constant as it is used as solvent and present in large amount. Therefore, the reaction is only first order with respect to sucrose. The hydrolysis of ester in presence of acid is first order reaction (keeping catalyst constant) CH3COOC2H5 + H2O [H ] + ⎯ →⎯ CH3COOH + C2H5OH Since [H2O] remain constant as in case of inversion of cane sugar, it does not effect the rate of reaction and reaction is simply first order with respect to ester. However, the hydrolysis of ester in presence of alkali CH2COOC2H5 + NaOH → CH3COONa + C2H5OH is second order being first order with respect to both ester and NaOH. While the molecularity of the reaction in each case, i.e. in hydrolysis of ester in presence of acid as well as in presence of alkali, is two. The reactions, in which molecularity and order are different due to the presence of one of the reactant in excess, are known as pseudo-order reactions. The word (pseudo) is always followed by order. For example, inversion of cane sugar is pseudo-first order reaction. The molecularity will always be a whole integer while order may be an integer, fraction or even a negative number. Molecularity is a theoretical concept, whereas order is empirical. Molecularity is, therefore, less significant as far as kinetic studies are concerned. The order of reaction provides the basis for classifying reactions. Generally, the order of reaction can be anywhere between zero and three. Reactions having order three and above are very rare and can be easily counted
6ChemicalKineticsandReactionDynamicsTherate of achemical reaction isproportionalto thenumberof collisionstakingplacebetweenthereactingmolecules andthechancesof simultaneouscollision of reacting molecules will go on decreasing with an increase innumber of molecules.Thepossibilityoffourormoremolecules comingcloser and colliding with one another at the same time is much less than incase of tri- or bi molecular reactions.Therefore, the reactions having orderfourormorearepracticallyimpossible.Further,manyreactionswhichappeartobequitecomplexproceedinstepwisechangesinvolvingmaximumtwoorthree species.The stoichiometric representation has no relation either withthemechanism of reaction or with the observed order of reaction.In olderliteraturethetermsunimolecular, bimolecularand termolecularhavebeen used to indicatethenumber of moleculesinvolved in asimplecollision process and should not be confused with first, second and thirdorderreactions1.4RateEquationsFor a reactionnAkProductThe rate is related with concentration of A withthefollowing differentialformofequationdAl = K[A]"(1.13)Rate =dtd[A]]or= log k + n log[A](1.14)dtwherek is therateconstant.As discussed previously the rate is determined by drawing a graph betweenconcentration and time andtaking the slope correspondingtoa concentration.If wehave thevalues oftheratesfor various concentrations,wecanfind theorderofreactionbyplotting log(rate)against log[concentration].The slopeof the straight lineobtainedfrom theplot gives the order of reactionn whilethe intercept gives log k.Thus, order and rate constant can be determined.However,theaveragerates calculatedbyconcentration versus timeplotsare not accurate.Even the values obtained as instantaneous rates by drawingtangents are subject to much error.Therefore, this method is not suitableforthe determination of order of a reaction as well as the value of the rateconstant. It is best tofind a method where concentration and time can besubstituted directly to determine the reaction orders.This could be achievedby integrating the differential rate equation.1.4.1 Integral Equations for nthOrder Reaction of a Single ReactantLet us consider the following general reaction:
6 Chemical Kinetics and Reaction Dynamics The rate of a chemical reaction is proportional to the number of collisions taking place between the reacting molecules and the chances of simultaneous collision of reacting molecules will go on decreasing with an increase in number of molecules. The possibility of four or more molecules coming closer and colliding with one another at the same time is much less than in case of tri- or bi molecular reactions. Therefore, the reactions having order four or more are practically impossible. Further, many reactions which appear to be quite complex proceed in stepwise changes involving maximum two or three species. The stoichiometric representation has no relation either with the mechanism of reaction or with the observed order of reaction. In older literature the terms unimolecular, bimolecular and termolecular have been used to indicate the number of molecules involved in a simple collision process and should not be confused with first, second and third order reactions. 1.4 Rate Equations For a reaction nA → k Product The rate is related with concentration of A with the following differential form of equation Rate = – [A] = [A] d dt k n (1.13) or log – [A] = log + log [A] d dt k n ⎛ ⎝ ⎞ ⎠ (1.14) where k is the rate constant. As discussed previously the rate is determined by drawing a graph between concentration and time and taking the slope corresponding to a concentration. If we have the values of the rates for various concentrations, we can find the order of reaction by plotting log (rate) against log [concentration]. The slope of the straight line obtained from the plot gives the order of reaction n while the intercept gives log k. Thus, order and rate constant can be determined. However, the average rates calculated by concentration versus time plots are not accurate. Even the values obtained as instantaneous rates by drawing tangents are subject to much error. Therefore, this method is not suitable for the determination of order of a reaction as well as the value of the rate constant. It is best to find a method where concentration and time can be substituted directly to determine the reaction orders. This could be achieved by integrating the differential rate equation. 1.4.1 Integral Equations for nth Order Reaction of a Single Reactant Let us consider the following general reaction:
Elementary1nA>ProductIf co is the initial concentration of the reactant and cthe concentration ofreactantatanytimetthedifferential rateexpressionmaybegivenas-dcldt = kcn(1.15)Multiplying by dt and then dividing by c",weget- dclc" = kdt(1.16)which may be integrated.The limits of integration are taken as c = Co and cat t = O and t = t, respectively, asdc-k(1.17)一Forvarious valuesof n,theresultsmaybe obtained as followsn=0;k=Co-c1n =l; In c =ln (co)-kt or c= Co e-ktn = 2; k = 1/t [1/c - 1/co]n=3;k=1/2t[1/c2_1/c]n = n; k= 1/(n-1)t [1/(cr-l) -1/(co-1)]1.4.2IntegralEquationsforReactions InvolvingMorethanOneReactantsWhen the concentrations of several reactants, and perhaps also products,appear in the rate expressions,it is more convenientto use as the dependentvariable x, i.e.the decrease in concentration of reactant in time t.Thenc =a-x, where a is commonly used to indicate the initial concentration inplaceof coand rateequation(1.15)becomesdx/dt = k (a-x)(1.18)[dx/(α -x)" = [kdtorwhich can be integrated taking the conditions: at t= O,x will also be zero,thevalueofrateconstantcanbeobtained.For various values of n the results obtained are as follows:n=0dx/dt=k; k=x/tn=1dxldt=k(a-x); k=2.303/tlogala-xn=2 dx/dt=k(a-x)2; k= 1/t[1/a-x-1/a]
Elementary 7 nA → k Product If c0 is the initial concentration of the reactant and c the concentration of reactant at any time t, the differential rate expression may be given as – dc/dt = kcn (1.15) Multiplying by dt and then dividing by cn , we get – dc/c n = kdt (1.16) which may be integrated. The limits of integration are taken as c = c0 and c at t = 0 and t = t, respectively, as ∫ ∫ dc c k dt n = (1.17) For various values of n, the results may be obtained as follows: n = 0; k C c t = – 0 n = 1; ln c = ln (c0) – kt or c = c0 e–kt n = 2; k = 1/t [1/c – 1/c0] n = 3; k = 1/2t [1/c2 – 1/ c0 2 ] n = n; k = 1/(n–1)t [1/(cn–1) – 1/( c n 0 –1 )] 1.4.2 Integral Equations for Reactions Involving More than One Reactants When the concentrations of several reactants, and perhaps also products, appear in the rate expressions, it is more convenient to use as the dependent variable x, i.e. the decrease in concentration of reactant in time t. Then c = a – x, where a is commonly used to indicate the initial concentration in place of c0 and rate equation (1.15) becomes dx/dt = k (a – x) n (1.18) or ∫ ∫ dx a x kdt n /( – ) = which can be integrated taking the conditions: at t = 0 , x will also be zero, the value of rate constant can be obtained. For various values of n the results obtained are as follows: n = 0 dx/dt = k; k = x/t n = 1 dx/dt = k(a – x); k = 2.303/t log a/a – x n = 2 dx/dt = k(a – x) 2 ; k = 1/t [1/a – x – 1/a]
8ChemicalKineticsandReactionDynamicsn = 3 dx/dt = k(a -x)3; k = 1/2t [1/(a -x)2- 1/a]n = n dx/dt = k(a-x); k= 1/(n - 1)t [1/(a-x)n-1_ 1/d-l]; n ≥21.5Half-lifeofaReactionThereactionrates can also be expressed in terms of half-lifeorhalf-lifeperiod.The half-life period is defined as the time required fortheconcentrationofareactanttodecreasetohalfof itsinitial value.Hence, half-life is the time required for one-half of the reaction to becompleted.It is represented by ti/2 and can be calculated by taking t= t1/2when x = al2 in the integrated rate equation of its order.Problem 1.1 Write the differential rate equations of the following reactions:(a)A+2B>P(b)3A+2Bk>3C+D+2ESolution The differential rates of above reactions can be written assumingthemtobeelementarystepsd[A]1 d[B]d[P] = k[A]? [B](a)dt2dtdt1 d[A]1 d[B]dD] - E = k [AP[B]?d[C]1(b)332dt2dtdtdtdtProblem1.2Write the differential rate equations of the following reactions:(a) A+3B→4C(b) A + 2B→C +3D(c) 3A +B+2C-→D+3ESolution Assuming these reactions as elementary steps,the differential ratecan be written as:d[A]1 d[B]I d[C]= k[A][B]3(a)dt34dtdtd[A]1 d[B]d[C]1 d[D]= k[A][B]?(b)2dtdtdtdtd[B]1 d[A]1 dIC _ dD] - ↓ d[E] = k[AP [B[C)(c)3dt2dtdtdt3dtProblem1.3Express therate constantk in unit of dm"mol-'s-1.if(i) k = 2.50× 10-9cm3molecule-'s-1(i) k = 2 × 10-° s-l atm-
8 Chemical Kinetics and Reaction Dynamics n = 3 dx/dt = k(a – x) 3 ; k = 1/2t [1/(a – x) 2 – 1/a2 ] n = n dx/dt = k(a – x) 3 ; k = 1/(n – 1)t [1/(a – x) n–1– 1/an–1]; n ≥ 2 1.5 Half-life of a Reaction The reaction rates can also be expressed in terms of half-life or half-life period. The half-life period is defined as the time required for the concentration of a reactant to decrease to half of its initial value. Hence, half-life is the time required for one-half of the reaction to be completed. It is represented by t1/2 and can be calculated by taking t = t1/2 when x = a/2 in the integrated rate equation of its order. Problem 1.1 Write the differential rate equations of the following reactions: (a) A + 2B → k P (b) 3A + 2B → k ′ 3C + D + 2E Solution The differential rates of above reactions can be written assuming them to be elementary steps (a) – [A] = – 1 2 [B] = [P] = [A] [B] 2 d dt d dt d dt k (b) – 1 3 [A] = – 1 2 [B] = 1 3 [C] = [D] = 1 2 [ ] = [A] [B] 3 2 d dt d dt d dt d dt d E dt k ′ Problem 1.2 Write the differential rate equations of the following reactions: (a) A + 3B → 4C (b) A + 2B → C + 3D (c) 3A + B + 2C → D + 3E Solution Assuming these reactions as elementary steps, the differential rate can be written as: (a) – [A] = – 1 3 [B] = 1 4 [C] = [A][B]3 d dt d dt d dt k (b) – [A] = – 1 2 [B] = [C] = 1 3 [D] = [A][B]2 d dt d dt d dt d dt k (c) – 1 3 [A] = – [B] = – 1 2 [C] = [D] = 1 3 [ ] = [A] [B][C] 3 2 d dt d dt d dt d dt d E dt k Problem 1.3 Express the rate constant k in unit of dm3 mol–1s –1, if (i) k = 2.50 × 10–9 cm3 molecule–1s –1 (ii) k = 2 × 10–6 s –1 atm–1