Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 14 Chemical Kinetics: The Rates of Chemical Reactions - Study Questions - Page 553g: 67


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Work Step by Step

The slow step is rate-determining so: $r=k_S[O_3][O]$ we need to eliminate the intermediary; the back fast reaction has the rate: $r_{-F}=k_{-F}[O_2][O]$ since it's an equilibrium $r_{-F}=r_F=k_{F}[O_3]$ and $K_F=k_F/k_{-F}$ $r=k_S[O_3]\dfrac{r_F}{k_{-F}[O_2]}=(K_Fk_S)[O_3]^2/[O_2]$
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