The activation energy of combined reactions

1. The activation energy of combined reactions • Consider that each of the rate constants of the following reactions A + A → A* + A (E1) A + A* → A + A (E1’) A* → P (E2) has an Arrhenius-like temperature dependence, one gets Thus the composite rate constant also has an Arrhenius-like form with activation energy, E = E1 + E2 – E1’ Whether or not the composite rate constant will increase with temperature depends on the value of E, if E > 0, k will increase with the increase of temperature

2. Combined activation energy

3. Theoretical problem 22.20 The reaction mechanism A2↔ A + A (fast) A + B → P (slow) Involves an intermediate A. Deduce the rate law for the reaction. • Solution:

4. Chain reactions • Chain reactions: a reaction intermediate produced in one step generates an intermediate in a subsequent step, then that intermediate generates another intermediate, and so on. • Chain carriers: the intermediates in a chain reaction. It could be radicals (species with unpaired electrons), ions, etc. • Initiation step: • Propagation steps: • Termination steps:

5. 23.1 The rate laws of chain reactions • Consider the thermal decomposition of acetaldehyde CH3CHO(g) → CH4(g) + CO(g) v = k[CH3CHO]3/2 it indeed goes through the following steps: 1. Initiation: CH3CHO → .CH3 + .CHO ki v = ki [CH3CHO] 2. Propagation: CH3CHO + .CH3 → CH4 + CH3CO.kp Propagation: CH3CO. → .CH3 + CO k’p 3. Termination: .CH3 + .CH3 → CH3CH3kt • The net rates of change of the intermediates are:

6. Applying the steady state approximation: • Sum of the above two equations equals: • thus the steady state concentration of [.CH3] is: • The rate of formation of CH4 can now be expressed as the above result is in agreement with the three-halves order observed experimentally.

7. Example: The hydrogen-bromine reaction has a complicated rate law rather than the second order reaction as anticipated. H2(g) + Br2(g) → 2HBr(g) Yield The following mechanism has been proposed to account for the above rate law. 1. Initiation: Br2 + M → Br. + Br. + M ki 2. Propagation: Br. + H2 → HBr + H.kp1 H. + Br2 → HBr + Br.kp2 3. Retardation: H. + HBr → H2 + Br.kr 4. Termination: Br. + Br. + M → Br2 + M* kt derive the rate law based on the above mechanism.

8. The net rates of formation of the two intermediates are • The steady-state concentrations of the above two intermediates can be obtained by solving the following two equations: • substitute the above results to the rate law of [HBr]

9. continued • The above results has the same form as the empirical rate law, and the two empirical rate constants can be identified as • Effects of HBr, H2, and Br2 on the reaction rate based on the equation

10. Self-test 23.1 Deduce the rate law for the production of HBr when the initiation step is the photolysis, or light-induced decomposition, of Br2 into two bromine atoms, Br.. Let the photolysis rate be v = Iabs, where Iabs is the intensity of absorbed radiation. • Hint: the initiation rate of Br. ?

11. Exercises 23.1b: On the basis of the following proposed mechanism, account for the experimental fact that the rate law for the decomposition 2N2O5(g) → 4NO2(g) + O2(g) is v = k[N2O5]. • N2O5 ↔ NO2 + NO3 k1, k1’ • NO2 + NO3 → NO2 + O2 + NO k2 • NO + N2O5 → NO2 + NO2 + NO2 k3

12. 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

13. The explosion limits of the H2 + O2 reaction

14. 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 intermediates 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

15. 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!!!

16. 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