This question is about reaction rates.
The reaction between peroxodisulfate(VI) ions, S2O82−(aq)\text{S}_2\text{O}_8^{2-}(\text{aq})S2O82−(aq), and iodide ions, I−(aq)\text{I}^-(\text{aq})I−(aq), in aqueous solution is shown below:
S2O82−(aq)+2I−(aq)→2SO42−(aq)+I2(aq)
\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})
S2O82−(aq)+2I−(aq)→2SO42−(aq)+I2(aq)
A student conducts three separate experiments at a constant temperature to investigate how varying the initial concentrations of the reactants affects the initial rate of this reaction. The experimental data obtained are presented below:
- Experiment 1:
- [S2O82−(aq)]=4.00×10−2 mol dm−3[\text{S}_2\text{O}_8^{2-}(\text{aq})] = 4.00 \times 10^{-2}\text{ mol dm}^{-3}[S2O82−(aq)]=4.00×10−2 mol dm−3
- [I−(aq)]=3.00×10−2 mol dm−3[\text{I}^-(\text{aq})] = 3.00 \times 10^{-2}\text{ mol dm}^{-3}[I−(aq)]=3.00×10−2 mol dm−3
- Initial rate=1.44×10−4 mol dm−3 s−1\text{Initial rate} = 1.44 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}Initial rate=1.44×10−4 mol dm−3 s−1
- Experiment 2:
- [S2O82−(aq)]=8.00×10−2 mol dm−3[\text{S}_2\text{O}_8^{2-}(\text{aq})] = 8.00 \times 10^{-2}\text{ mol dm}^{-3}[S2O82−(aq)]=8.00×10−2 mol dm−3
- [I−(aq)]=3.00×10−2 mol dm−3[\text{I}^-(\text{aq})] = 3.00 \times 10^{-2}\text{ mol dm}^{-3}[I−(aq)]=3.00×10−2 mol dm−3
- Initial rate=2.88×10−4 mol dm−3 s−1\text{Initial rate} = 2.88 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}Initial rate=2.88×10−4 mol dm−3 s−1
- Experiment 3:
- [S2O82−(aq)]=4.00×10−2 mol dm−3[\text{S}_2\text{O}_8^{2-}(\text{aq})] = 4.00 \times 10^{-2}\text{ mol dm}^{-3}[S2O82−(aq)]=4.00×10−2 mol dm−3
- [I−(aq)]=6.00×10−2 mol dm−3[\text{I}^-(\text{aq})] = 6.00 \times 10^{-2}\text{ mol dm}^{-3}[I−(aq)]=6.00×10−2 mol dm−3
- Initial rate=2.88×10−4 mol dm−3 s−1\text{Initial rate} = 2.88 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}Initial rate=2.88×10−4 mol dm−3 s−1
