23.2 Explosions Thermal explosion: a very rapid reaction arising from a rapid increase of reaction rate with increasing temperature. Chain-branching explosion: occurs when the number of chain centres grows exponentially. An example of both types of explosion is the following reaction 2H2(g) + O2(g) → 2H2O(g) 1. Initiation: H2 → H. + H. 2. Propagation H2 + .OH → H. + H2O kp 3. Branching: O2 + .H → O + .OH kb1 O + H2 → .OH + H. Kb2 4. Termination H. + Wall → ½ H2 kt1 H. + O2 + M → HO2. + M* kt2

The explosion limits of the H2 + O2 reaction

Analyzing the reaction of hydrogen and oxygen (see preceding slide), show that an explosion occurs when the rate of chain branching exceeds that of chain termination. Method: 1. Set up the corresponding rate laws for the reaction intermediate and then apply the steady-state approximation. 2. Identify the rapid increase in the concentration of H. atoms. Applying the steady-state approximation to .OH and O gives

Therefore, we write kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M], then At low O2 concentrations, termination dominates branching, so kterm > kbranch. Then this solution corresponds to steady combustion of hydrogen. At high O2 concentrations, branching dominates termination, kbranch > kterm. Then This is an explosive increase in the concentration of radicals!!!

Self-test 23.2 Calculate the variation in radical composition when rates of branching and termination are equal. Solution: kbranch = 2kb1[O2] and kterm = kt1 + kt2[O2][M], The integrated solution is [H.] = vinit t

Polymerization kinetics Stepwise polymerization: any two monomers present in the reaction mixture can link together at any time. The growth of the polymer is not confined to chains that are already formed. Chain polymerization: an activated monomer attacks another monomer, links to it, then that unit attacks another monomer, and so on.

23.3 Stepwise polymerization Commonly proceeds through a condensation reaction, in which a small molecule is eliminated in each step. The formation of nylon-66 H2N(CH2)6NH2 + HOOC(CH2)4COOH → H2N(CH2)6NHOC(CH2)4COOH HO-M-COOH + HO-M-COOH → HO-M-COO-M-COOH Because the condensation reaction can occur between molecules containing any number of monomer units, chains of many different lengths can grow in the reaction mixture.

Stepwise polymerization The rate law can be expressed as Assuming that the rate constant k is independent of the chain length, then k remains constant throughout the reaction. The degree of polymerization: The average number of monomers per polymer molecule, <n>

23.4 Chain polymerization Occurs by addition of monomers to a growing polymer, often by a radical chain process. Rapid growth of an individual polymer chain for each activated monomer. The addition polymerizations of ethene, methyl methacrylate, and styrene. The rate of polymerization is proportional to the square root of the initiator concentration.

The three basic types of reaction step in a chain polymerization Initiation: I → R. + R. vi = ki[I] M + R. → .M1 (fast) (b) Propagation: M + .M1→ .M2 M + .M2→ .M3 ░ vp = kp[M][.M] M + .Mn-1→ .Mn (c) Termination: Mutual termination: .Mn + .Mm→ Mn+m Disproportionation: .Mn + .Mm→ Mn + Mm Chain transfer: M + .Mn→ Mn + .M

Influences of termination step on the polymerization Mutual termination: two growing radical chains combine. vt = kt ([.M])2 Disproportionation: Such as the transfer of a hydrogen atom from one chain to another, which corresponds to the oxidation of the donor and the reduction of acceptor. vt = kt ([.M])2 Chain transfer: vt = ?

the net rate of change of radical concentration is calculated as Using steady-state approximation (the rate of production of radicals equals the termination rate) The rate of polymerization vp = kp[.M][M] = kp[M] The above equation states that the rate of polymerization is proportional to the square root of the concentration of the initiator. Kinetic chain length, v, <n> = 2v (for mutual termination)

(a) The steady-state concentration of free radicals. Example: For a free radical addition polymerization with ki = 5.0x10-5 s-1 , f = 0.5, kt = 2.0 x107 dm3 mol-1 s-1, and kp = 2640 dm3 mol-1 s-1 , and with initial concentrations of [M] = 2.0 M and [I] = 8x10-3 M. Assume the termination is by combination. (a) The steady-state concentration of free radicals. (b) The average kinetic chain length. (c) The production rate of polymer. Solution: (a) (b) (c) The production rate of polymer corresponds to the rate of polymerization is vp: vp = kp[.M][M]

23.5 Features of homogeneous catalysis A Catalyst is a substance that accelerates a reaction but undergoes no net chemical change. Enzymes are biological catalysts and are very specific. Homogeneous catalyst: a catalyst in the same phase as the reaction mixture. heterogeneous catalysts: a catalyst exists in a different phase from the reaction mixture.

Example: Bromide-catalyzed decomposition of hydrogen peroxide: 2H2O2(aq) → 2H2O(l) + O2(g) is believed to proceed through the following pre-equilibrium: H3O+ + H2O2 ↔ H3O2+ + H2O H3O2+ + Br- → HOBr + H2O v = k[H3O2+][Br-] HOBr + H2O2 → H3O+ + O2 + Br- (fast) The second step is the rate-determining step. Thus the production rate of O2 can be expressed by the rate of the second step. The concentration of [H3O2+] can be solved [H3O2+] = K[H2O2][H3O+] Thus The rate depends on the concentration of Br- and on the pH of the solution (i.e. [H3O+]).

Exercise 23.4b: Consider the acid-catalysed reaction (1) HA + H+ ↔ HAH+ k1, k1’ , both fast (2) HAH+ + B → BH+ + AH k2, slow Deduce the rate law and show that it can be made independent of the specific term [H+] Solution: