Question

The reduction of peroxodisulphate ion by I\(^{-}\) ion is expressed by \(\text{S}_2\text{O}_8^{2-} + 3\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_3^-\). If rate of disappearance of I\(^{-}\) is \(9/2 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\), what is the rate of formation of \(2\text{SO}_4^{2-}\) during same time?

• A. \(3 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
• B. \(2 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
• C. \(10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)
• D. \(4 \times 10^{-3}\,\text{mol L}^{-1}\text{s}^{-1}\)

Hint:

Use stoichiometric coefficients to relate reaction rates.

Answer 1

The Correct Option is A

Step 1: Stoichiometric ratio: \[ 3\,\text{I}^- \rightarrow 2\,\text{SO}_4^{2-} \]
Step 2: \[ \text{Rate of SO}_4^{2-} = \frac{2}{3} \times \frac{9}{2}\times10^{-3} = 3\times10^{-3} \]