The decomposition of ozone, , follows the mechanism: Step 1: (fast equilibrium, forward, backward); — Chemical Kinetics Chemistry Question
Question
The decomposition of ozone, $2O_3 \rightarrow 3O_2$, follows the mechanism: Step 1: $O_3 \rightleftharpoons O_2 + O$ (fast equilibrium, $k_1$ forward, $k_{-1}$ backward); Step 2: $O + O_3 \rightarrow 2O_2$ (slow, $k_2$). Using the rapid equilibrium assumption, the correct effective rate law is:
Answer: A
💡 Solution & Explanation
The rate determining step is the slow step: $\text{Rate} = k_2[O][O_3]$. Since atomic oxygen ($O$) is an intermediate, we apply equilibrium to the fast step: $K_{eq} = \frac{k_1}{k_{-1}} = \frac{[O_2][O]}{[O_3]} \Rightarrow [O] = \frac{k_1 [O_3]}{k_{-1} [O_2]}$. Substituting this intermediate concentration into the RDS gives: $\text{Rate} = k_2 \left( \frac{k_1 [O_3]}{k_{-1} [O_2]} \right) [O_3] = \frac{k_1 k_2}{k_{-1}} [O_3]^2 [O_2]^{-1}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes