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26 Nov 2019

the following mechanisim for decompostion of ozone in theatmosphere

O3-----> O2+O and its reverse(k1,kz')

O+ O3----> O2+O2 (k2, the reserve rxn is nigligibly slow)

use steadt-state with O treated as an intermediate, to find therate of decompostion of O3, show that step 2 is slow, then the rateis second order in O3 and -1 in O2

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The decomposition of ozone into oxygen is given by the followingreaction equation:

2 O3 (g) ?----> 3 O2(g)

The proposed mechanism is as follows:

(1) O3 ? O2+O k1

(2) O + O2 ?O3 k2

(3) O + O3 ?2O2 k3

(a) Derive an expression for the reaction rate, R, as afunction of the rate constants and concentrations of oxygen andozone. You may apply the steady state approximation for the O atomintermediate.

(b) Under what conditions is the reaction first order inozone?

PLEASE EXPLAIN ALL OF THIS BETTER THAN THE BOOK AND DO EXAMPLE 14.9!!!

Ozone naturally decomposes to oxygen by this reaction: 2 O_3 (g) rightarrow 3 O_2 (g) The experimentally observed rate law for this reaction is: Rate = k [O_3]^2 [O_2]^-1 Show that this proposed mechanism is consistent with the experimentally observed rate law. O_3 (g) O_2 (g) + O (g) Fast O_3 (g) + O(g) 2 O_2 (g) Slow To determine whether the mechanism is valid, you must First determine whether the steps sum to the overall reaction. The steps do indeed sum to the overall reaction, so the first condition is met. O_3 (g) O_2 (g) + O (g) O_3 (g) + O (g) 2 O_2 (g)/2 O_3 (g) rightarrow 3 O_2 (g) The second condition is that the rate law predicted by the mechanism is consistent with the experimentally observed rate law. Because the second step is rate limiting, write the rate law based on the second step. Rate = k_2 [O_3] [O] Because the rate law contains an intermediate (O), you must express the concentration of the intermediate in terms of the concentrations of the reactants of the overall reaction. To do this, set the rates of the forward reaction and the reverse reaction of the first step equal to each other. Solve the expression from the previous step for [O], the concentration of the intermediate. Rate (forward) = Rate (backward) k_1 [O_3] = k_-1 [O_2] [O] [O] = k_1 [O_3]/k_-1 [O_2] Finally, substitute [O] into the rate law predicted by the slow step. Rate = k_2 [O_3] [O] = k_2 [O_3] k_1 [O_3]/k_-1 [O_2] = k_2 k_1 [O_3]^2/k_-1 [O_2] = k [O_3]^2 [O_2]^-1 CHECK Because the two steps in the proposed mechanism sum to the overall reaction, and because the rate law obtained from the proposed mechanism is consistent with the experimentally observed rate law, the proposed mechanism is valid. The -1 reaction order with respect to [02J indicates that the rate slows down as the concentration of oxygen increases- oxygen inhibits, or slows down, the reaction. FOR PRACTICE 14.9 Predict the overall reaction and rate law that results from the following two-step mechanism: 2 A rightarrow A_2 Slow A_2 + B rightarrow A_2 B Fast

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