23.2 Explosions

1. 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 H2kt1 H. + O2 + M → HO2. + M* kt2

2. The explosion limits of the H2 + O2 reaction

3. 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

4. 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!!!

5. 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.] = vinitt

6. 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.

7. 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.

8. 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>

9. 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.

10. 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

11. 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 = ?

12. 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)

13. 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]

14. 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.

15. 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+]).

16. 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: