Question

The equation for the reaction between iodide and bromate ions in acidic solution is \(6 \mathrm{I}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 3 \mathrm{I}_{2}(a q)+\mathrm{Br}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}\) The rate of the reaction is followed by measuring the appearance of \(\mathrm{I}_{2}\). The following data are obtained: $$ \begin{array}{clcc} \hline\left[I^{-}\right] & {\left[\mathrm{BrO}_{3}^{-}\right]} & {\left[\mathrm{H}^{+}\right]} & \text {Initial } \operatorname{Rate}(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.0020 & 0.0080 & 0.020 & 8.89 \times 10^{-5} \\ 0.0040 & 0.0080 & 0.020 & 1.78 \times 10^{-4} \\ 0.0020 & 0.0160 & 0.020 & 1.78 \times 10^{-4} \\ 0.0020 & 0.0080 & 0.040 & 3.56 \times 10^{-4} \\ 0.0015 & 0.0040 & 0.030 & 7.51 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the hydrogen ion concentration when the rate is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and \(\left[\mathrm{I}^{-}\right]=\left[\mathrm{BrO}_{3}^{-}\right]=0.0075 \mathrm{M} ?\)

Short answer

The hydrogen ion concentration under these conditions is \(0.178 \ \mathrm{M}\).

Step-by-step solution

Determine the order with respect to iodide ions

Observe how the rate changes when the concentration of \(\mathrm{I}^{-}\) is changed while keeping the concentrations of \(\mathrm{BrO}_{3}^{-}\) and \(\mathrm{H}^{+}\) constant. Use the first and second rows of data: Initial rate when \([\mathrm{I}^{-}]\) is doubled: \(\frac{1.78 \times 10^{-4}}{8.89 \times 10^{-5}} = 2\) Since the rate doubled when the concentration of \(\mathrm{I}^{-}\) doubled, the order of the reaction with respect to \(\mathrm{I}^{-}\) is 1.

Determine the order with respect to bromate ions

Now, observe how the rate changes when the concentration of \(\mathrm{BrO}_{3}^{-}\) is changed while keeping the concentrations of \(\mathrm{I}^{-}\) and \(\mathrm{H}^{+}\) constant. Use the first and the third rows of data: Initial rate when \([\mathrm{BrO}_{3}^{-}]\) is doubled: \(\frac{1.78 \times 10^{-4}}{8.89 \times 10^{-5}} = 2\) Since the rate doubled when the concentration of \(\mathrm{BrO}_{3}^{-}\) doubled, the order of the reaction with respect to \(\mathrm{BrO}_{3}^{-}\) is 1.

Determine the order with respect to hydrogen ions

Now, observe how the rate changes when the concentration of \(\mathrm{H}^{+}\) is changed while keeping the concentrations of \(\mathrm{I}^{-}\) and \(\mathrm{BrO}_{3}^{-}\) constant. Use the first and the fourth rows of data: Initial rate when \([\mathrm{H}^{+}]\) is doubled: \(\frac{3.56 \times 10^{-4}}{8.89 \times 10^{-5}} = 4\) Since the rate quadrupled when the concentration of \(\mathrm{H}^{+}\) doubled, the order of the reaction with respect to \(\mathrm{H}^{+}\) is 2.

Write the rate expression for the reaction

Using the orders found in steps 1-3, the rate expression for the reaction is: \(rate = k[\mathrm{I}^{-}]^1[\mathrm{BrO}_{3}^{-}]^1[\mathrm{H}^{+}]^2\)

Calculate k

Use the rate expression formula and the data from the first row of the table to find the value of k: \(8.89 \times 10^{-5} \ \mathrm{mol / L \cdot s}= k(0.0020 \ \mathrm{M})^1(0.0080 \ \mathrm{M})^1(0.020 \ \mathrm{M})^2\) \(k = \frac{8.89 \times 10^{-5}}{(0.0020)(0.0080)(0.020)^2} = 278.125 \ \mathrm{M^{-2} \cdot s^{-1}}\)

Find the hydrogen ion concentration

With the given rate, iodide ion concentration, and bromate ion concentration, we can find the hydrogen ion concentration using the rate expression and the calculated value of k: \(5.00 \times 10^{-4} \ \mathrm{mol / L \cdot s} = 278.125 \ \mathrm{M^{-2} \cdot s^{-1}}(0.0075 \ \mathrm{M})(0.0075 \ \mathrm{M})([\mathrm{H}^{+}]^2)\) \([\mathrm{H}^{+}]^2 = \frac{5.00 \times 10^{-4}}{278.125(0.0075)^2} = 0.0316\) \([\mathrm{H}^{+}] = \sqrt{0.0316} = 0.178 \ \mathrm{M}\) So, the hydrogen ion concentration when the rate is \(5.00 \times 10^{-4} \ \mathrm{~mol / L \cdot s}\) and \([\mathrm{I}^{-}] = [\mathrm{BrO}_{3}^{-}] = 0.0075 \ \mathrm{M}\) is \(0.178 \ \mathrm{M}\).