Solution:
The rate of formation of the brown-coloured tri-iodide ion ΔI3−/Δt\Delta \text{I}_3^-/\Delta tΔI3−/Δt can be determined by measuring the rate of appearance of the colour. Given the initial rate data at 25°C, we can analyze how the concentration of H2O2\text{H}_2\text{O}_2H2O2 and I−\text{I}^-I− affect the rate.
From the data:
- Experiment 1: [H2O2]=0.100[ \text{H}_2\text{O}_2 ] = 0.100[H2O2]=0.100 M, [I−]=0.100[ \text{I}^- ] = 0.100[I−]=0.100 M, Initial rate = 1.15×10−41.15 \times 10^{-4}1.15×10−4 M/s
- Experiment 2: [H2O2]=0.100[ \text{H}_2\text{O}_2 ] = 0.100[H2O2]=0.100 M, [I−]=0.200[ \text{I}^- ] = 0.200[I−]=0.200 M, Initial rate = 2.30×10−42.30 \times 10^{-4}2.30×10−4 M/s
- Experiment 3: [H2O2]=0.200[ \text{H}_2\text{O}_2 ] = 0.200[H2O2]=0.200 M, [I−]=0.100[ \text{I}^- ] = 0.100[I−]=0.100 M, Initial rate = 2.30×10−42.30 \times 10^{-4}2.30×10−4 M/s
- Experiment 4: [H2O2]=0.200[ \text{H}_2\text{O}_2 ] = 0.200[H2O2]=0.200 M, [I−]=0.200[ \text{I}^- ] = 0.200[I−]=0.200 M, Initial rate = 4.60×10−44.60 \times 10^{-4}4.60×10−4 M/s
From these experiments, it's observed that doubling the concentration of either H2O2\text{H}_2\text{O}_2H2O2 or I−\text{I}^-I− doubles the initial rate of formation of I3−\text{I}_3^-I3−. Therefore, the reaction is first order with respect to both H2O2\text{H}_2\text{O}_2H2O2 and I−\text{I}^-I−.