Question

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+$$\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results:

Short answer

The given reaction is: $2Cl{O}_2(aq)+2O{H}^{-}(aq)⟶Cl{O}_3^{-}(aq)+Cl{O}_2^{-}(aq)+H_{2}O(l)$. Based on the provided hypothetical experimental data, the reaction is found to be first order with respect to ClO₂ and zero order with respect to OH⁻. The rate law for the reaction is: Rate = k [ClO₂], where k is the rate constant.

Step-by-step solution

Determine the reaction order with respect to ClO₂ and OH⁻.

To determine the reaction order with respect to ClO₂, compare trials 1 and 2, where the concentration of OH⁻ remains constant. By doubling the concentration of ClO₂, the initial rate also doubles. This implies that the reaction is first order with respect to ClO₂. Now, compare trials 2 and 3, where the concentration of ClO₂ remains constant, and the concentration of OH⁻ is halved. The initial rate remains the same, suggesting that the reaction is zero order with respect to OH⁻.

Write the rate law for the reaction.

From the reaction orders determined in the previous step, we can write the rate law for the reaction as follows: Rate = k [ClO₂]¹[OH⁻]⁰ Since the reaction is zero order with respect to OH⁻, the rate law simplifies to: Rate = k [ClO₂] The reaction rate depends only on the concentration of ClO₂ and the rate constant, k.