Physical chemistr Reaction Kinetics(3) Xuan Cheng Xiamen University
1 Reaction Kinetics (3) Xuan Cheng Xiamen University Physical Chemistry
Ch ical chemistr Reaction Kinetic Determination of the rate law 1. Half-life method The rate law r=kA] [B].m (1748) /2 Forn≠1 (n-DA1- (17.29 2 og1041/2=g10 (n-1)logIoLAlo (17.49 (n-1) log101/2 g10 1-n)logIo[A] (1749)
2 Determination of the Rate Law Physical Chemistry The rate law r = k[A] [B] [L] (17.48) 1. Half-life method o A n n A n k t ( 1)log [ ] ( 1) 2 1 log log 10 1 10 1/ 2 10 − − − − = − (17.49) A n o n n A k t 1 1 1/ 2 ( 1) 2 1 − − − − = For n 1 (17.29) o A n n A n k t (1 )log [ ] ( 1) 2 1 log log 10 1 10 1/ 2 10 + − − − = − (17.49) Reaction Kinetics
Ch ical chemistr Reaction Kinetic 半衰期法确定反应级数 用半衰期法求除一级反应以外的其它反应的级数。 根据n级反应的半衰期通式:2联两个不同起始 浓度1A7o,4作实验,分别测定半衰期为1和 因過同一反应,常数相同,所以 h(1/2/1/2) n=1+ 1/2 In(Alo/Alo) h/2=hK+(1-m)n 以nt1~ln47作图从直线斜率求n值。从多个实验数据 用作图法求出的n值更加准确
3 半衰期法确定反应级数 用半衰期法求除一级反应以外的其它反应的级数。 以lnt1/2～ln[A]o作图从直线斜率求n值。从多个实验数据 用作图法求出的n值更加准确。 根据 n 级反应的半衰期通式： 取两个不同起始 浓度[A]o，[A]o ’作实验，分别测定半衰期为t1/2和 ， 因为同一反应，常数相同，所以： 1 1 1/ 2 [ ] 1 ( 1) 2 1 − − − − = n A o n n k A t 1/ 2 t ' Physical Chemistry 1 1/ 2 1/ 2 [ ] [ ] ' ' − = n o o A A t t ln([ ] '/[ ] ) ln( / ' ) 1 1/ 2 1/ 2 A o A o t t n = + A o ln t ln K (1 n)ln[ ] 1/ 2 = + − Reaction Kinetics
Ch ical chemistr Reaction Kinetic Determination of the rate law 2. Powell-plot method r=k[4 C≡ O p≡kA[A (17.50) the fraction of a unreacted +[B-(n-1) kgt For n≠1(1728 k Forn=1(17.13 C Forn≠ In a=-o For n=1 (17.51)
4 Physical Chemistry Determination of the Rate Law 2. Powell-plot method A A o [ ]/[ ] n r = k[A] (17.50) the fraction of A unreacted k A t n A o 1 [ ] − A n k t A A A n o n o 1 ( 1) 1 1 = + − − − For n 1 (17.28) 1 ( 1) 1 − = − − n n For n 1 (17.13) k t A A A o ln = − For n = 1 ln = − For n = 1 (17.51) Reaction Kinetics
Ch ical chemistr Reaction Kinetic Determination of the rate law Forn≠1 In a For (17.51) For a given n, there is a fixed relation between a and o for every reaction of order n Plot a versus logroo for commonly occurring values of n to give a series of master curves. (Fig. 17.6) The Powell-plot method requires the initial investment of time needed to make the master plots Table 17.1 gives the data needed to make the master plots
5 Physical Chemistry Determination of the Rate Law 1 ( 1) 1 − = − − n n For n 1 ln = − For n = 1 (17.51) For a given n, there is a fixed relation between and for every reaction of order n. Plot versus log10 for commonly occurring values of n to give a series of master curves. (Fig. 17.6) The Powell-plot method requires the initial investment of time needed to make the master plots. Table 17.1 gives the data needed to make the master plots. Reaction Kinetics
Ch ical chemistr Reaction Kinetic Determination of the rate law 3. Initial-rate method The rate law r=k42[B/…L 1748) Measure ro for two different initial concentrations LAo. and LAJo.2 while keeping B o,/C],.fixed The ratio of initial rates for runs 1 and 2 0.2 702/70= a can be found The orders B,...1 can be Io ound similarly
6 Physical Chemistry Determination of the Rate Law 3. Initial-rate method The rate law r = k[A] [B] [L] (17.48) The ratio of initial rates for runs 1 and 2 = 0,1 0,2 0,2 0,1 [ ] [ ] / A A r r Measure r0 for two different initial concentrations [A]0,1 and [A]0,2 while keeping [B]0 , [C]0 , …fixed. can be found The orders ,… can be found similarly Reaction Kinetics
Ch ical chemistr Reaction Kinetic Determination of the rate law Isolation method The rate law r=kA[BI (1748) Make initial concentrations of reactant a much less than the concentrations of all other species [B]o>>lA C0>>lalo The rate law becomes r=kA[B.LL=jIAja where j=[B]-[L1( 17.52 Where i is essentially constant The reaction has the pseudo-order a The orders B,.n can be found similarly
7 Physical Chemistry Determination of the Rate Law 4. Isolation method The rate law r = k[A] [B] [L] (17.48) Make initial concentrations of reactant A much less than the concentrations of all other species [B]0 >> [A]0, [C]0 >> [A]0, … The rate law becomes [ ] [ ] [ ] [ ] r = k A B 0 L 0 = j A (17.52) 0 0 where j k[B] [L] Where j is essentially constant. The reaction has the pseudo-order . The orders ,… can be found similarly. Reaction Kinetics
Ch ical chemistr Reaction Kinetic 孤立法确定反应级数 孤立法类似于准级数法,它不能用来确定反应级数, 而只能使问题简化,然后用前面三种方法来确定反应 级数 r=k[A][B] 1使A]>B]r=kB]先确定 2使[B>[A]r=k"[A]再确定Q值
8 孤立法确定反应级数 孤立法类似于准级数法，它不能用来确定反应级数， 而只能使问题简化，然后用前面三种方法来确定反应 级数。 r = k[A] [B] 1.使[A]>>[B] r = k'[B] 先确定β值 2.使[B]>>[A] r = k''[A] 再确定α值 Physical Chemistry Reaction Kinetics
Ch ical chemistr Reaction Kinetic Rate laws and equilibrium Constants for Elementary reactions Show that for a reaction that takes place in a sequence of steps, the overall equilibrium constant is a product of ratios of the rate constants for each step It is sufficient to consider a reasonably general but simple two-step reaction sequence, such as A+BOC+d(second-order in each direction, k k)) COE+F (first-order forwarded, second-order reverse, k,, k. A+B分D+E+F( overall) d)=-14081+k1(CD1(40)=-k2(]+k1E门1
9 Physical Chemistry Rate Laws and Equilibrium Constants for Elementary Reactions A+ B C+ D [ ][ ] [ ][ ] [ ] k1 A B k 1 C D dt d A = − + − Show that for a reaction that takes place in a sequence of steps, the overall equilibrium constant is a product of ratios of the rate constants for each step. It is sufficient to consider a reasonably general but simple two-step reaction sequence, such as C E+ F A+ B D+ E+ F (second-order in each direction, k1 , k-1 ) (first-order forwarded, second-order reverse, k2 , k-2 ) (overall) [ ] [ ][ ] [ ] k2 C k 2 E F dt d C = − + − Reaction Kinetics
Ch ical chemistr Reaction Kinetic Rate laws and equilibrium Constants for Elementary reactions At equilibrium, all the reaction are individually at equilibrium, and setting the net rates each equal to zero gives -kilaB+kcd dcl [C]+k-2EIF] [CD] K1 EFI k LaCk The equilibrium constant of the overall reaction is therefore KD=CD月2CD[=Axk2 [AJB [AB[C [A[B [C]k_1k-2 elementary reaction (17.53)
10 Physical Chemistry Rate Laws and Equilibrium Constants for Elementary Reactions At equilibrium, all the reaction are individually at equilibrium, and setting the net rates each equal to zero gives 1 1 [ ][ ] [ ][ ] − = k k A B C D 2 2 [ ] [ ][ ] − = k k C E F [ ][ ] [ ][ ] [ ] k1 A B k 1 C D dt d A = − + − [ ] [ ][ ] [ ] k2 C k 2 E F dt d C = − + − 2 2 1 1 [ ] [ ][ ] [ ][ ] [ ][ ] [ ][ ][ ] [ ][ ][ ][ ] [ ][ ] [ ][ ][ ] − − = = = = k k k k C E F A B C D A B C C D E F A B D E F Kc The equilibrium constant of the overall reaction is therefore b f c k k K = elementary reaction (17.53)* Reaction Kinetics
