CHM 311 –CHEMICAL KINETICS MODULE 2A Dr. C. J. Ajaelu REACTION MECHANISM, STEADY STATE APPROXIMATION, PRE-EQUILIBRIA Learning

MODULE 2A

Dr. C. J. Ajaelu

REACTION MECHANISM, STEADY STATE APPROXIMATION, PRE-EQUILIBRIA

Learning objective:

 To understand the meaning of reaction mechanism  To understand elementary reactions and molecularity  To understand and explain rate determining step  To understand and explain steady state approximation  To derive a general expression of the rate law using the steady state approximation  To understand and explain pre-equilibria  To derive a general expression of the rate law using the pre- equilibria

Reaction mechanism This refers to the sequence of elementary steps that take place from reactants to the formation of product. Each step in a reaction mechanism is called the elementary reaction, elementary step or elementary process. Elementary Reaction involves molecules or ions in a reaction mechanism.

These elementary steps occur precisely as represented in the step equation and they add up to yield the overall equation. Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Consider the reactions

O3(g) O2(g) + O Step 1 (slow)

O + O3(g) 2O2(g) Step 2 (fast)

2O3(g) 3O2(g) Overall reaction

Step 1 and step 2 are elementary reactions. Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates. The slowest step in a reaction mechanism is the rate determining step. It is the step that controls the overall rate of the reaction.

Molecularity of a reaction

This is the number of reactant species (molecules, ions atoms) involved in a chemical reaction in an elementary step. Unimolecular reaction involves reaction of a single molecule to form product. This single molecule can form a new arrangement as in the isomerization of cyclopropane into propene. Another example of unimolecular reaction is the gas phase decomposition of cyclobutane,

C4H10 to ethylene, C2H2.

C4H8 2C2H2

d[C H ] The rate law for the above unimolecular reaction is given by 4 8  k[C H ] dt 4 8

Thus the rate law for a unimolecular elementary reaction is first order in the reactant.

d[X ] X Z  k[X ] dt

Bimolecular reaction involves two reacting molecules colliding with each other and exchanging energy, atoms, or groups of atoms, or undergo some other kind of change to form product(s). Thus the rate law for a bimolecular elementary reaction is second order in the reactants. d[X ] X + Y Z  k[X ][Y] dt

The difference between order of a reaction and molecularity of a reaction is that order of a reaction is an empirical quantity, and can be derived from the rate law experimentally while molecularity is an elementary reaction proposed as an individual step in a mechanism. Termolecular reactions: These are reactions that involve three reacting molecules.

Steady State approximation

Reaction mechanisms involving many steps are difficult to solve analytically, hence an alternative method of solution is necessary. One method is to integrate the rate law numerically. A more widely used method is to make an approximation. An approximation, called the steady-state approximation (also called the quasi-steady state approximation) assumes that after an initial induction period, an interval during which the concentrations of intermediates, I, rise from zero, and during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are negligibly small. Simply put, the steady state approximation assumes that the concentration of reaction intermediates remains constant throughout the reaction. It is also assumed that the concentration of the intermediate is consumed as fast as it is produced. Using consecutive reaction below to explain the steady state: X I Z

The rate constant for the formation of I is k1 while the rate constant for the formation of Z is k2.

1.8

k1[X] - k2[I] = 0 1.9 k1[X] is the rate of formation of the intermediate I and it is positive because I is formed. K2[I] is the rate of consumption of the intermediate I and it is negative because I is consumed. Eq. 1.9 deals with the rate of formation and consumption of the intermediate I. From equation 1.9, make [I] the subject of the formula k [I]  1 [X ] 1.10 k2 though [X] does depend on the time, the dependence of [I] on the time is negligible k For eq. 1.10 to be consistent with eq. 1.8, 1 ˂˂1 k2

Z is formed by a first-order decay of X, with a rate constant k1, the rate constant of the slower, rate-determining step. We can write down the solution of this equation at once by substituting the solution for [X], and integrating d[Z]  k [I]  k [X ] 1.11 dt 2 1 The first order decay of X is given by

k1t [X ]  [X ]o e 1.12 From eqs. 1.11 and 1.12

t d[Z]  k [X ] ek1tdt 1 o  1.13 0

k1t = (1 e )[X]o 1.14

Example

Derive the rate law for the chain reaction for the formation of methane.

CH3 CHO CH4 + CO

The proposed mechanism for the reaction is given by

CH3CHO CH4 + CO k1 CH3CHO + CH3 CH4 + CH3CO k2

CH3CO CH3 + CO k3

CH3 + CH3 C2H6 k4

Solution

1st identify the intermediates

Secondly, write expressions for the rate of formation of the intermediates and set them equal to zero. d[CH ] 3  k [CH CHO] k [CH CHO][CH ] k [CH CO] k [CH ]2  0 dt 1 3 2 3 3 3 3 4 3 1.15 d[CH 3CO]  k2[CH 3CHO][CH 3 ] k3[CH 3CO]  0 dt 1.16

Solving these two equations simultaneously (-k2[CH3CHO][CH3] and k2[CH3CHO][CH3] cancel out, + k3[CH3CO] and -k3[CH3CO] cancel out), we have

2 k1[CH3CHO] =k4[CH3]

1 2  k  1  1  2 [CH 3 ]    [CH 3CHO] 1.17  k4 

The overall rate of reaction can be obtained by the formation of CH4 in step 4. The rate equation is given by

d[CH ] 1.18 4  k [CH CHO][CH ]  0 dt 2 3 3

Substitute for [CH3] in eq. 1.17 into eq. 1.18

1 d[CH ]  k  2 1 4  1  2  k2[CH 3CHO]  [CH 3CHO] 1.19 dt  k2 

1 d[CH ]  k  2 3 4  k  1  [CH CHO]2 dt 2  k  3  4  1.20

Note that it is the intermediates ([CH3] and [CH3CHO]) you approximate to zero.

CH4 is not an intermediate, hence it is not approximated or equated to zero.

Pre-equilibria

Consider a consecutive reaction in which reactants form an intermediate I and the intermediate finally forms a product. The intermediate formed is at equilibrium with the reactants X and Y X + Y I Z 1.21

The rate constants are k1 and k-1 for the forward and reverse reactions of the equilibrium respectively, and k2 for the final step. Equation 1.21 involves a pre- equilibrium, in which an intermediate is in equilibrium with the reactants. A pre- equilibrium can arise when the rate at which the intermediate decays to form the reactants is much faster than the rate at which it forms products; thus, the condition is possible when k-1 >> k2 but not when k2 >> k-1. There is an assumption that X, Y and Z are in equilibrium, thus the equilibrium constant K can be written as

k where K  1 1.22

k1 Equation 1.22 presumes that the rate of formation of the product Z from the intermediate I is too slow to affect the maintenance of the pre-equilibrium. Thus, the rate of formation of Z may now be written: dZ  k [I]  k K[X ][Y] 1.23 dt 2 2

The rate has a form of a second order and its rate constant is a composite one

Thus, the rate law can be re- written as

dZ k1k2  k[X ][Y] where k = k2K = 1.24 dt k1