The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate $=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )$

The balanced chemical equation for the reaction between hydrogen selenite and iodate can be written as follows: \[ 2\mathrm{IO_3}^- + 5\mathrm{H_2SeO_3} + 12\mathrm{H^+} \rightarrow I_3^- + 5\mathrm{Se} + 6\mathrm{H_2O} \] Step 2: Determine the order of the reaction with respect to each reactant

To determine the order of the reaction with respect to \(\mathrm{IO_3^-}\), we will analyze the experimental data provided. We can see that when the concentration of \(\mathrm{IO_3^-}\) is doubled, the reaction rate doubles as well. This indicates that the reaction is first-order with respect to \(\mathrm{IO_3^-}\). The rate expression for \(\mathrm{IO_3^-}\) is given by: \[ Rate = k[\mathrm{IO_3^-}]^n \] Now, let's analyze the experimental data provided for \(\mathrm{H_2SeO_3}\). We can see that when the concentration of \(\mathrm{H_2SeO_3}\) is doubled, the reaction rate remains the same. This indicates that the reaction is zero-order with respect to \(\mathrm{H_2SeO_3}\). Thus, the rate law can be expressed as: \[ Rate = k[\mathrm{IO_3^-}]^n[\mathrm{H_2SeO_3}]^0 \] Step 3: Calculate the rate constant

Given that rate \(= -\frac{\Delta[\mathrm{H_2SeO_3}]}{\Delta t}\). We can use the experimental data and the rate expression we found in step 2 to calculate the rate constant. Let's use the data from the first row of the table: \(Rate = 4.0 \times 10^{-4} \frac{mol}{L \cdot min}\) \[\left[\mathrm{IO_3^-}\right] = 2.0 \times 10^{-3} \frac{mol}{L}\] Using the rate expression: \[Rate = k[\mathrm{IO_3^-}]^n\] We can plug in the values and find the rate constant: \[4.0\times10^{-4} = k(2.0\times10^{-3})^n\] Solving for k, we get: \[k = \frac{4.0\times 10^{-4}}{(2.0\times 10^{-3})}\] Thus, the rate constant, k, equals \(2 \times 10^{-1} \frac{L}{mol\cdot min}\). Step 4: Write the final rate law expression

Combining our findings from Steps 2 and 3, we can now write the complete rate law for the reaction: \[Rate = (2 \times 10^{-1} \frac{L}{mol\cdot min})[\mathrm{IO_3^-}]^1[\mathrm{H_2SeO_3}]^0\] Or simply as: \[Rate = 2 \times 10^{-1}[\mathrm{IO_3^-}]\] So, the rate law is \(Rate = 2 \times 10^{-1}[\mathrm{IO_3^-}]\) and the value of the rate constant is \(2 \times 10^{-1} \frac{L}{mol\cdot min}\).