The activation energy of combined reactions

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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 (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

Combined activation energy

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:

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:

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 → . CH CHO ki v = ki [CH3CHO] 2. Propagation: CH3CHO + . CH3 → CH4 + CH3CO kp Propagation: CH3CO. → .CH CO k’p 3. Termination: .CH CH3 → CH3CH3 kt The net rates of change of the intermediates are:

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.

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: Br M → Br Br. + M ki 2. Propagation: Br H2 → HBr + H kp1 H Br2 → HBr Br kp2 3. Retardation: H HBr → H Br kr 4. Termination: Br Br. + M → Br M* kt derive the rate law based on the above mechanism.

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]

Effects of HBr, H2, and Br2 on the reaction rate based on the equation
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

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. ?

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 + NO k1, k1’ NO NO3 → NO2 + O2 + NO k2 NO + N2O5 → NO2 + NO2 + NO2 k3

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 H OH → H H2O kp 3. Branching: O H → .O OH kb1 .O H2 → .OH H Kb2 4. Termination H. + Wall → ½ H2 kt1 H. + O M → HO 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 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

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

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