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The activation energy of combined reactions Consider that each of the rate constants of the following reactions A + A → A* + A (E 1 ) A + A* → A + A (E.

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Presentation on theme: "The activation energy of combined reactions Consider that each of the rate constants of the following reactions A + A → A* + A (E 1 ) A + A* → A + A (E."— Presentation transcript:

1 The activation energy of combined reactions Consider that each of the rate constants of the following reactions A + A → A* + A (E 1 ) A + A* → A + A (E 1 ’) A* → P (E 2 ) 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 The reaction mechanism A 2 ↔ 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 CH 3 CHO(g)→CH 4 (g)+ CO(g) v = k[CH 3 CHO] 3/2 it indeed goes through the following steps: 1. Initiation: CH 3 CHO→. CH 3 +. CHO k i v = k i [CH 3 CHO] 2. Propagation: CH 3 CHO +. CH3 → CH 4 + CH 3 CO. k p Propagation: CH3CO. →. CH3 + CO k’ p 3. Termination:. CH3 +. CH3 → CH 3 CH 3 k t 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 CH 4 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. H 2 (g) + Br 2 (g) → 2HBr(g) Yield The following mechanism has been proposed to account for the above rate law. 1. Initiation: Br 2 + M → Br. + Br. + M k i 2. Propagation: Br. + H 2 → HBr + H. k p1 H. + Br 2 → HBr + Br. k p2 3. Retardation: H. + HBr → H 2 + Br. k r 4. Termination: Br. + Br. + M → Br 2 + M* k t 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, H 2, and Br 2 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 Br 2 into two bromine atoms, Br.. Let the photolysis rate be v = I abs, where I abs 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 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) is v = k[N 2 O 5 ]. (1)N 2 O 5 ↔ NO 2 + NO 3 k 1, k 1 ’ (2)NO 2 + NO 3 → NO 2 + O 2 + NO k 2 (3)NO + N 2 O 5 → NO 2 + NO 2 + NO 2 k 3

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 2H 2 (g) + O 2 (g) → 2H 2 O(g) 1. Initiation:H 2 → H. + H. 2. PropagationH 2 +. OH → H. + H 2 Ok p 3. Branching:O 2 +. H →. O +. OHk b1. O + H 2 →. OH + H. K b2 4. TerminationH. + Wall → ½ H 2 k t1 H. + O 2 + M → HO 2. + M* k t2

13 The explosion limits of the H 2 + O 2 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 k branch = 2k b1 [O 2 ] and k term = k t1 + k t2 [O 2 ][M], then At low O 2 concentrations, termination dominates branching, so k term > k branch. Then this solution corresponds to steady combustion of hydrogen. At high O 2 concentrations, branching dominates termination, k branch > k term. 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: k branch = 2k b1 [O 2 ] and k term = k t1 + k t2 [O 2 ][M], The integrated solution is [H. ] = v init t


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