Given the rate data below from a series of kinetics experiments, determine the orders for the following reaction, and state the overall order of the reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+3 \mathrm{I}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{cccc} \text { Experiment }\left[\mathbf{H}_{2} \mathrm{O}_{2}\right] & {\left[\mathbf{I}^{-}\right]} & {\left[\mathbf{H}^{+}\right]} & \text {Rate }(\mathbf{M} / \mathbf{s}) \\ \hline 1 & 0.010 \mathrm{M} & 0.010 \mathrm{M} & 0.00050 \mathrm{M} & 1.15 \times 10^{-6} \\ 2 & 0.020 \mathrm{M} & 0.010 \mathrm{M} & 0.00050 \mathrm{M} & 2.30 \times 10^{-6} \\ 3 & 0.010 \mathrm{M} & 0.020 \mathrm{M} & 0.00050 \mathrm{M} & 2.30 \times 10^{-6} \\ 4 & 0.010 \mathrm{M} & 0.010 \mathrm{M} & 0.00100 \mathrm{M} & 1.15 \times 10^{-6} \end{array}$$

Step by step solution

01

Compare rates of Experiment 1 and Experiment 2

Here, the concentration of I- and H+ are unchanged, allowing us to find the order with respect to H2O2: \(\frac{Rate_2}{Rate_1} = \frac{k[\mathrm{H}_{2}\mathrm{O}_{2}]_2^x[\mathrm{I}^-]_2^y[\mathrm{H}^+]_2^z}{k[\mathrm{H}_{2}\mathrm{O}_{2}]_1^x[\mathrm{I}^-]_1^y[\mathrm{H}^+]_1^z}\) Step 2: Solve for x

02

Calculate reaction order for H2O2

Plugging in the values from the table, we get: \(\frac{2.30\times10^{-6}}{1.15\times10^{-6}} = \frac{[\mathrm{H}_{2}\mathrm{O}_{2}]_2^x}{[\mathrm{H}_{2}\mathrm{O}_{2}]_1^x}\) Solving for x, we find x = 1. Step 3: Pick two experiment pairs for I-

03

Compare rates of Experiment 1 and Experiment 3

Here, the concentration of H2O2 and H+ are unchanged, allowing us to find the order with respect to I-: \(\frac{Rate_3}{Rate_1} = \frac{k[\mathrm{H}_{2}\mathrm{O}_{2}]_3^x[\mathrm{I}^-]_3^y[\mathrm{H}^+]_3^z}{k[\mathrm{H}_{2}\mathrm{O}_{2}]_1^x[\mathrm{I}^-]_1^y[\mathrm{H}^+]_1^z}\) Step 4: Solve for y

04

Calculate reaction order for I-

Plugging in the values from the table, we get: \(\frac{2.30\times10^{-6}}{1.15\times10^{-6}} = \frac{[\mathrm{I}^-]_3^y}{[\mathrm{I}^-]_1^y}\) Solving for y, we find y = 1. Step 5: Pick two experiment pairs for H+

05

Compare rates of Experiment 1 and Experiment 4

Here, the concentration of H2O2 and I- are unchanged, allowing us to verify the order with respect to H+: \(\frac{Rate_4}{Rate_1} = \frac{k[\mathrm{H}_{2}\mathrm{O}_{2}]_4^x[\mathrm{I}^-]_4^y[\mathrm{H}^+]_4^z}{k[\mathrm{H}_{2}\mathrm{O}_{2}]_1^x[\mathrm{I}^-]_1^y[\mathrm{H}^+]_1^z}\) Step 6: Solve for z

06

Verify reaction order for H+

Plugging in the values from the table, we get: \(\frac{1.15\times10^{-6}}{1.15\times10^{-6}} = \frac{[\mathrm{H}^+]_4^z}{[\mathrm{H}^+]_1^z}\) Since the rates are equal, z must be 0. Step 7: Calculate the overall order

07

Determine the overall order of the reaction

Overall order is the sum of x, y, and z: Overall order = 1 + 1 + 0 = 2

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