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: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{CIO}_{2}\right](M)} & {\left[\mathrm{OH}^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ 2 & 0.020 & 0.030 & 0.00276 \\ 3 & 0.020 & 0.090 & 0.00828 \\ \hline \end{array} $$ (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

From the given table, we can compare the rates of the different experiments to determine the order of the reaction with respect to each reactant. Consider experiments 1 and 2. The [ClO2] in experiment 1 is 3 times that of experiment 2, while [OH-] remains the same. Comparing the rates of these two experiments, we get: \(\frac{0.0248}{0.00276} = 9\) Since the [ClO2] increased by a factor of 3, and the rate increased by a factor of 9, we can deduce that the order of the reaction with respect to ClO2 is 2 (first order: rate increase would be 3; second order: rate increase would be \(3^2 = 9\); etc.) Now, let's consider experiments 2 and 3. The [OH-] in experiment 3 is 3 times that of experiment 2 while the [ClO2] remains the same. Comparing the rates of these two experiments, we get: \(\frac{0.00828}{0.00276} = 3\) Since the [OH-] increased by a factor of 3, and the rate increased by a factor of 3 as well, we can deduce that the order of the reaction with respect to OH- is 1 (first-order reaction since rate increase is same as concentration increase). Thus, the rate law for the reaction can be written as: Rate = k[ClO2]^2[OH-]^1

Now that we have determined the rate law, we can calculate the rate constant 'k' using the concentration and rate data from any of the experiments. We will use the data from experiment 1. Rate = k[ClO2]^2[OH-]^1 0.0248 = k(0.060)^2(0.030) Solving for k: k = \(\frac{0.0248}{(0.060)^2(0.030)}\) = 9.2 M^(-2)s^(-1) The rate constant (k) is 9.2 M^(-2)s^(-1).

Now that we have the rate law and the rate constant, we can calculate the rate of the reaction when [ClO2] = 0.100 M and [OH-] = 0.050 M. Rate = k[ClO2]^2[OH-]^1 Rate = (9.2 M^(-2)s^(-1))(0.100 M)^2(0.050 M) Rate = 0.046 M/s The rate of the reaction when [ClO2] = 0.100 M and [OH-] = 0.050 M is 0.046 M/s.