State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Statistical Thermodynamics and Chemical Kinetics State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Lecture 9 May 12, 2003

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Chapter 9 Complex Reactions 9.1 Exact analytic solutions for complex reactions Introduction Most chemical processes are complex, i.e. they consist of a number of coupled elementary reactions. These complex reactions can be divided into several classes: (1) opposing or reversible reactions, (2) consecutive reactions, (3) parallel reactions, and (4) mixed reactions. In this chapter, we examine methods for determining exact analytic solutions for the time dependence of concentrations of species involved in complex reactions.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Reversible reactions In reversible reactions or opposing reactions, the products of the initial reaction can proceed to re-form the original substances. A chemical example of this is the cis-trans isomerization of 1,2-dichloroethylene: (9-1) As such, the simplest reversible reaction is of the form (9-2) and is first order in each direction. The differential equation for this mechanism is (9-3) (9-4)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 If it is assumed that both A 1 and A 2 are present in the system at time t=0, that is, [A 1 ]=[A 1 ] 0 and [A 2 ]=[A 2 ] 0, then at any time afterwords the total amount of reactant remaining and new product formed must equal the initial amount of the reactants before reactions. Hence, [A 1 ] 0 + [A 2 ] 0 = [A 1 ] + [A 2 ] (9-5) Solving for [A 2 ], we obtain [A 2 ] = [A 1 ] 0 + [A 2 ] 0 - [A 1 ] (9-6) and substituting this into equation (9-3) yields (9-7)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 To find the solution of equation (9-7) we introduce a variable m, defined as (9-8) This allows us to rewrite equation (9-7) as (9-9) which we may then integrate (9-10) The solution is (9-11)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 If only [A 1 ] is present in the system initially, at t=0, then the solution reduces to (9-12) which is just (9-13) Using the mass conservation constraint [A 1 ] 0 =[A 1 ] + [A 2 ], we can obtain the solution for [A 2 ] : (9-14) When equilibrium is reached, the individual reactions must be balanced; in other words, the reaction A  B must occur just as frequently as the reverse reaction. The forward and reverse reactions occur at the same rate.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Consequently, for reaction (9-2) at equilibrium, (9-15) and (9-16) and we have the following definition of the equilibrium constant K eq expressed in terms of rate constants: (9-17) The same argument can be extended to a reversible reaction that occurs in multiple stages.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First-order reversible reactions involving two steps Reversible reactions may be distinguished by number of stages and the number of initial reactants involved in the reaction. Here, we consider the complete derivation for first order reversible reactions involving only two stages, that is, (9-18) The kinetic equations for the system are (9-19) (9-20) (9-21)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 We assume that at t=0, [A 1 ]=[A 1 ] 0, and [A 2 ] 0 =[A 3 ] 0 =0 and that the amounts of A 1, A 2 and A 3 which have reacted at a later time satisfy the equation: [A 1 ] 0 =[A 1 ] + [A 2 ] + [A 3 ] (9-22) By the principle of detailed balance, we have and (9-23,24) Using equations (9-22)-(9-24) gives (9-25) So (9-26)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Substituting equation (9-26) into equation (9-19) we obtain the new first-order differential equaion: (9-27) Solving this euation using standard methods, we have (9-28) Substituting equation (9-22),(9-26) and (9-28) into this expression, we have, upon simplification, (9-29)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 As t  , (9-30) (9-31) (9-32) and the system is then in a state of equilibrium. The misuse of the principle arises when the inter- mediate A 2 is difficult to detect, so that an experimentalist might think that A 1  A 3 is an elementary reaction. Thus the equilibrium constant for [A 3 ] e /[A 1 ] e might be taken to be k 1 /k -2. However, since A 1  A 3 is not an elementary reaction, this conclusion is incorrect.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 To determine K = [A 3 ] e /[A 1 ] e correctly, the equilibrium of each elementary reaction must be considered, that is, equations (9-23) and (9-24) must be used. The correct expression of K can be found: (9-33) Another example of a first-order reversible reaction involving only two stages is the cyclic reaction

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First and second order reversible reactions. A reversible reaction may be of mixed order, such as, An example is N 2 O 4  2NO 2. The rate expression for this type of reaction is (9-34) To find the solution, we introduce a progress variable, x = [A 1 ] 0 - [A 1 ] (9-35) Equation (9-34) thus becomes

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Or (9-36) We let (9-37) and So that (9-38) Then (9-39) The solution of equation(9-39) is (9-40) Where

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Consecutive Reactions Irreversible reactions can be defined as those which start with an initial reactant and produce products or intermediates generally in only one direction. Consecutive reactions are sequential irreversible reactions. There are two classes of consecutive reactions: Those which are first order and those which are mixed first order and second order.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First-order consecutive reactions First-order with two steps. Consider the consecutive first-order reaction involving two stages: (9-42) This mechanism can be described by the following set of rate expressions: (9-43) (9-44) (9-45)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 The concentration of A 1 is obtained after integration as (9-46) The concentration of A 2 is computed from equation (9-44) (9-47) Solving this equation, we obtain the time dependence of [A 2 ]: (9-48) If [A 2 ] 0 =0 at t=0, then (9-49)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 The concentration of A 3 can be determined from conservation of mass: [A 1 ] 0 = [A 1 ] + [A 2 ] + [A 3 ] (9-50) Hence we have (9-51) Simplifying this expression gives (9-52)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First-order with three steps. The system of differential equations for a first-order consecutive reaction involving three steps, viz.: (9-53) can be integrated in a way similar to the two-step case. The differential equations for A 1 and A 2 and their solutions do not differ from those obtained in the two-step case, while the differential equation for A 3 is (9-54) Substituting equation (9-49) for [A 2 ] into equation (9-54) gives (9-55)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Integrating this linear equation subject to the initial condition that [A 3 ] =0 at t=0, we obtain (9-56) Then (9-57)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Higher order consecutive reactions. Some higher order consecutive reactions may be first order in one step and second order the in second, such as (9-58) Or both steps may be second order consecutive reactions, as in (9-59) The differential equations for most of these systems of reactions are nonlinear and generally they have no exact solutions. Nevertheless, analytic solutions for such systems can be obtained if time is eliminated as a variable. To illustrate this method, consider the reaction sequence described by the system of equations (2-58):

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 (9-60) (9-61) (9-62) To solve this system of equations we divide equation (9- 61) by (9-60) (9-63) If we let K=k 1 /k 2, then (9-64)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Equation (9-64) can be rearranged to (9-65) (9-66) Integrating equation(9-66) over the respective limits, i.e., (9-67) gives (9-68)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Parallel reactions Parallel reactions are defined as two or more processes in which the same species participate in each reaction step. The most common cases of parallel reactions are: (1) those in which the initial reactant decomposes into several different products; (2) those in which the initial reactants are different, but yield the same products; and (3) those in which a substance reacts with two or more initial reactants.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First order decay to different products Consider the mechanism (9-69) At t=0 the initial concentrations of the four components are [A 1 ]= [A 1 ] 0 ; [A 2 ]= [A 3 ] = [A 4 ] = 0 (9-70,71) The differential equation for the system can be written, as exemplified for [A 1 ], (9-72) Integrating equation (9-72) yields (9-73)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 To solve for [A 2 ] we substitute equation (9-73) into equation (9-74), (9-74) And upon integrating we obtain (9-75) Similarly we obtain (9-76) (9-77)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 It follows that the relative rate constants can be determined by measuring the relative product yields: (9-78) This ratio defines the branching ratio for the reaction; note that this branching ration is independent of time. The above example can be easily generalized to reactions of a single reactant into n different products. We thus have (9-79) (9-80)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 First order decay to the same products. Consider the reaction sequence (9-81) The rate expressions for the disappearance of A 1 and A 3 and the appearance of A 2 are (9-82) (9-83) (9-84) At t=0, [A 1 ]=[A 1 ] 0, [A 3 ]=[A 3 ] 0, and [A 2 ]=0.

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 The equations describing the time dependence for each component are (9-85) (9-86) Since (9-87) Upon integration we obtain (9-88) or (9-89)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Parallel second-order reactions In the case of parallel second-order reactions, (9-90) Two differential equations can be written, (9-91) (9-92) Conservation of mass demands [A 4 ] = [A 2 ] 0 - [A 2 ] (9-93) [A 5 ] = [A 3 ] 0 - [A 3 ] (9-94) [A 1 ] 0 - [A 1 ] = [A 4 ] +[A 5 ]=[A 2 ] 0 -[A 2 ] + [A 3 ] 0 -[A 3 ] (9-95)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Eliminating time as a variable, yields (9-96) which can be integrated under the initial condition that [A 3 ]=[A 3 ] 0 and [A 2 ] = [A 2 ] 0 at time t=0. This gives (9-97) From the mass conservation relation we can determine [A 5 ] by substituting equation (9-94) into equation (9-97). Rearranging then gives (9-98)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 The expression for [A 1 ] is obtained from equation (9-95) [A 1 ] = [A 1 ] 0 - [A 2 ] 0 - [A 3 ] 0 +[A 2 ] + [A 3 ] (9-99) Substituting equation (9-96) into equation (9-99) gives (9-100) Using this expression for [A 1 ], we obtain a differential equation for [A 2 ]: (9-101) (9-102)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Integrating (9-102), we obtain (9-103) There is no explicit solution to this integral, but we can consider some limiting cases. Case I. k 1 >> k 2 Under this condition the ration k 2 /k 1 is approximately zero, thus equation (9-103) reduces to the simple form (9-104) (9-105)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Rearranging and solving for [A 2 ] t yields (9-106) Case II. k 1 = k 2 Under this condition the ration k 2 /k 1 =1, thus equation (9- 103) can be integrable, viz. (9-107) (9-108)

State Key Laboratory for Physical Chemistry of Solid Surfaces 厦门大学固体表面物理化学国家重点实验室 Assignments: 1. Consider the reaction at t=0, [A 1 ]=[A 1 ] 0, [A 2 ]=[A 3 ]=…= [A n-1 ]=[A n ]=0. Derive an expression that will describe the concentration of the intermediates existing at any time t during the reaction. [E. Abel, Z. Phys. Chem. A56, 558(1906).] 2.Consider the first-order cyclic reaction Assume that initially at t=0, [A]=[A] 0, [B]=[C]=0, and the amounts of A,B and C always follow [A] 0 =[A] + [B] + [C] Using the detailed balanced method, solve for [A],[B] and [C].