Question

For the reaction,
\( 3\,\text{I}^-_{(aq)} + \text{S}_2\text{O}_8^{2-}{}_{(aq)} \longrightarrow 2\,\text{SO}_4^{2-}{}_{(aq)} + \text{I}_3^-{}_{(aq)} \)
rate of formation of \( \text{SO}_4^{2-}{}_{(aq)} \) is \( 0.044\,\text{mol dm}^{-3}\,\text{s}^{-1} \).
What is the rate of consumption of \( \text{S}_2\text{O}_8^{2-}{}_{(aq)} \) ?

• A. \(0.022 \text{ mol dm}^{-3} \text{ s}^{-1}\)
• B. \(0.044 \text{ mol dm}^{-3} \text{ s}^{-1}\)
• C. \(0.066 \text{ mol dm}^{-3} \text{ s}^{-1}\)
• D. \(0.088 \text{ mol dm}^{-3} \text{ s}^{-1}\)

Hint:

Rate relation: Divide by coefficient of that species.

Answer 1

The Correct Option is A

Concept:
Rate is related to stoichiometric coefficients: \[ \frac{-1}{1}\frac{d[\text{S}_2\text{O}_8^{2-}]}{dt} = \frac{1}{2}\frac{d[\text{SO}_4^{2-}]}{dt} \]

Step 1: Given \[ \text{Rate of formation of } SO_4^{2-} = 0.044 \]

Step 2: Apply relation \[ \text{Rate of consumption of } S_2O_8^{2-} = \frac{1}{2} \times 0.044 = 0.022 \] Conclusion: \[ \text{Answer = 0.022 mol dm}^{-3}\text{s}^{-1} \]