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Chapter 12: Problem 102

The thiosulfate ion \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right)\) is oxidized by iodine as follows: $$ 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ In a certain experiment, \(7.05 \times 10^{-3}\) mol/L of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) . Calculate the rate of production of iodide ion.

Short Answer

Expert verified
The rate of consumption of the thiosulfate ion (S₂O₃²⁻) in the reaction is \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\), and the rate of production of iodide ion (I⁻) is also \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\), since they have a 1:1 stoichiometric relationship.

Step by step solution

01

Calculate the rate of consumption of S₂O₃²⁻

To calculate the rate of consumption of S₂O₃²⁻, we will use the formula: Rate of consumption = (Change in concentration) / (Change in time) We are given that the change in concentration of S₂O₃²⁻ is 7.05 × 10⁻³ mol/L, and the change in time is 11.0 seconds. Therefore, we can plug in these values into our formula: Rate of consumption of S₂O₃²⁻ = (7.05 × 10⁻³ mol/L) / (11.0 s)
02

Perform the calculation

Now we can perform the calculation to find the rate of consumption of S₂O₃²⁻: Rate of consumption of S₂O₃²⁻ = \( \frac{7.05 \times 10^{-3} \, mol/L} {11.0 \, s} \) = \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\)
03

Calculate the rate of production of I⁻

To calculate the rate of production of I⁻, we will use the stoichiometric relationship between S₂O₃²⁻ and I⁻ from the balanced chemical equation: 2 S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2 I⁻ From the balanced equation, we can see that for every 2 moles of S₂O₃²⁻ consumed, 2 moles of I⁻ are produced. Therefore, we have a 1:1 molar ratio between the consumption of S₂O₃²⁻ and the production of I⁻. This means that the rate of production of I⁻ will be equal to the rate of consumption of S₂O₃²⁻, which we calculated in Step 2. Rate of production of I⁻ = Rate of consumption of S₂O₃²⁻ = \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\)
04

Finalize the answer

The rate of consumption of the thiosulfate ion (S₂O₃²⁻) in the reaction is \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\), and the rate of production of iodide ion (I⁻) is also \(6.41 \times 10^{-4} \frac{mol}{L \cdot s}\), since they have a 1:1 stoichiometric relationship.

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The rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

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