Question

Consider the reaction,
\(3\text{I}_{(aq)}^-+\text{S}_2\text{O}_8^{2-}_{(aq)}⟶\text{I}_3^-_{(aq)}+2\text{SO}_4^{2-}_{(aq)}\)
If the rate of formation of \(\text{SO}_4^{2-}\) at a particular time is \(2.2\times 10^{-2}\,\text{mol dm}^{-3}\,\text{s}^{-1}\). Calculate the rate of consumption of \(\text{I}^-\).

• A. \(1.1\times 10^{-2}\,\text{mol dm}^{-3}\,\text{s}^{-1}\)
• B. \(2.2\times 10^{-2}\,\text{mol dm}^{-3}\,\text{s}^{-1}\)
• C. \(3.3\times 10^{-2}\,\text{mol dm}^{-3}\,\text{s}^{-1}\)
• D. \(4.4\times 10^{-2}\,\text{mol dm}^{-3}\,\text{s}^{-1}\)

Hint:

Divide each rate by its stoichiometric coefficient to compare rates.

Answer 1

The Correct Option is C

Step 1: Understand the concept
For \(a\text{A} \to b\text{B}\) the rate of reaction is \(-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}\). Each species rate is divided by its coefficient.

Step 2: Write the rate expression
For \(3\text{I}^- + \text{S}_2\text{O}_8^{2-} \to \text{I}_3^- + 2\text{SO}_4^{2-}\):
\[ -\frac{1}{3}\frac{d[\text{I}^-]}{dt} = \frac{1}{2}\frac{d[\text{SO}_4^{2-}]}{dt} \]

Step 3: Substitute
\[ -\frac{d[\text{I}^-]}{dt} = \frac{3}{2} \times 2.2 \times 10^{-2} = 3.3 \times 10^{-2}\ \text{mol dm}^{-3}\text{s}^{-1} \]

Step 4: Check options
2.2e-2 would be right only if iodide had the same coefficient as sulphate, and 1.1e-2 or 4.4e-2 come from mistaken ratios of 1/2 or 2. The true ratio is 3 : 2.

Final Answer:
Iodide is used at 3.3 x 10^-2 mol dm^-3 s^-1. This is option (C). \[ \boxed{\text{(C) }3.3 \times 10^{-2}\ \text{mol dm}^{-3}\text{s}^{-1}} \]