Answer to Question #241736 in General Chemistry for TUTU

Question

 A suggested mechanism for the decomposition of ozone is as follows: O3 O2 + O fast equilibrium O + O3 2O2 slow step What is the rate law predicted by this mechanism? 

Answer 1

For a reaction, mechanism follows the next elementary steps:

Here, intermediate is O as it forms in one elementary step and it is consumed in next steps.

Rate law comes from slowest step that is,

$rate = k_2[O][O_3] \;\;\;(1)$

For fast step at equilibrium, rate of forward and reverse reactions are equal.

$k_1[O_3] =k_{-1}[O_2][O] \\ [O] = \frac{k_1[O_3]}{k_{-1}[O_2]} \;\;\;(2)$

Substitute [O] in rate law (1).

$rate = k_2[O][O_3] \\ = k_2\frac{k_1[O_3]}{k_{-1}[O_2]}[O_3] \\ = \frac{k_1k_2[O_3]^2}{k_{-1}[O_2]}$