91Ó°ÊÓ

To determine the rate law equation, we need to find the relationship between the rate of the reaction (the rate of disappearance of peroxydisulfate) and the initial concentrations of the reactants. The general rate law equation is: \(rate = k[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m[\mathrm{I}^{-}]^n\) where \(k\) is the rate constant, \(m\) and \(n\) are the reaction orders with respect to the peroxydisulfate and iodide ions, respectively. To find the reaction orders (m and n), we can use the initial rate data from the table. Let's start with experiments 1 and 2 since they both have the same initial concentration of iodide ions: Experiments 1 and 2: \(\dfrac{rate_2}{rate_1} = \frac{3.9 \times 10^{-6}}{2.6 \times 10^{-6}} = 1.5\) \(\dfrac{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_2[\mathrm{I}^{-}]^n_2}{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_1[\mathrm{I}^{-}]^n_1}\) Since the concentration of iodide ions is the same, it will cancel out: \(\dfrac{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_2}{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_1} = 1.5 \Rightarrow \dfrac{(0.027)^m}{(0.018)^m} = 1.5\) After isolating m and solving, we find that \(m \approx 1\). Next, let's use the data from experiments 1 and 3 since they both have the same initial concentration of peroxydisulfate ions: Experiments 1 and 3: \(\dfrac{rate_3}{rate_1} = \frac{7.8 \times 10^{-6}}{2.6 \times 10^{-6}} = 3\) \(\dfrac{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_3[\mathrm{I}^{-}]^n_3}{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}]^m_1[\mathrm{I}^{-}]^n_1}\) Since the concentration of peroxydisulfate ions is the same, it will cancel out: \(\dfrac{[\mathrm{I}^{-}]^n_3}{[\mathrm{I}^{-}]^n_1} = 3 \Rightarrow \dfrac{(0.054)^n}{(0.036)^n} = 3\) After isolating n and solving, we find that \(n \approx 2\). Thus, the rate law equation is: \(rate = k[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}][\mathrm{I}^{-}]^2\)

Now, we can use any experiment's data along with the determined rate law to calculate the rate constant for that experiment. Then, we can do this for all experiments and calculate the average rate constant. For experiment 1: \(rate_1 = k[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}_1][\mathrm{I}^{-}_1]^2 \Rightarrow k = \dfrac{rate_1}{[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}_1][\mathrm{I}^{-}_1]^2} = 7.14 \times 10^3\, M^{-2}s^{-1}\) Similarly, for experiments 2, 3, and 4: For experiment 2: \(k = 5.96 \times 10^3\, M^{-2}s^{-1}\) For experiment 3: \(k = 5.71 \times 10^3\, M^{-2}s^{-1}\) For experiment 4: \(k = 5.56 \times 10^3\, M^{-2}s^{-1}\) Now, we can take the average of these values: \(k_{avg} = \dfrac{(7.14 + 5.96 + 5.71 + 5.56) \times 10^3}{4} \, M^{-2}s^{-1} = 6.09 \times 10^3\, M^{-2}s^{-1}\)

Using stoichiometry from the balanced equation, for every molecule of peroxydisulfate (\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)) that disappears, there are three molecules of iodide ion (\(\mathrm{I}^{-}\)) that also disappear. Thus, the rate of disappearance of the iodide ion (Rate\( _{\mathrm{I}^{-}}\)) is three times the rate of disappearance of the peroxydisulfate ion (Rate\(_{\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}}\)): \(Rate_{\mathrm{I}^{-}} = 3 \cdot Rate_{\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}}\)

We are given the initial concentrations of peroxydisulfate and iodide ions: \([\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}] = 0.025\, M\) and \([\mathrm{I}^{-}] = 0.050\, M\). We can calculate the rate of peroxydisulfate disappearance using the rate law equation and average rate constant: \(Rate_{\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}} = k_{avg}[\mathrm{S}_{2}\mathrm{O}_{8}{ }^{2-}][\mathrm{I}^{-}]^2 = 6.09 \times 10^3 \cdot 0.025 \cdot (0.050)^2 = 3.8 \times 10^{-6}\,M/s\) Now, we can calculate the rate of iodide ion disappearance using the relationship found in step 3: \(Rate_{\mathrm{I}^{-}} = 3 \cdot Rate_{\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}} = 3 \cdot (3.8 \times 10^{-6}) = 11.4 \times 10^{-6}\, M/s\)