Solution:
The rate of formation of the brown-coloured tri-iodide ion 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 and I− affect the rate.
From the data:
- Experiment 1: [H2O2]= 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] = 0.200[I−]=0.200 M, Initial rate = 2.30×10−42.30 \times 10^{-4}2.30×10−4 M/s
- Experiment 3: [H2O2]= 0.100[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[I−]=0.200 M, Initial rate = 4.60×10−44.60 \times 10^{-4}4.60×10−4 M/s
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