Question

For the reaction, $3\text{I}^-_{\text{(aq)}} + \text{S}_2\text{O}^{2-}_{8\text{(aq)}} \longrightarrow \text{I}^-_{3\text{(aq)}} + 2\text{SO}^{2-}_{4\text{(aq)}}$, rate of formation of $\text{SO}^{2-}_4$ is $0.022\ \text{mol dm}^{-3}\ \text{sec}^{-1}$. What is rate of formation of $\text{I}^-_{3\text{(aq)}}$?

• A. $0.022\ \text{mol dm}^{-3}\ \text{sec}^{-1}$
• B. $0.11\ \text{mol dm}^{-3}\ \text{sec}^{-1}$
• C. $0.011\ \text{mol dm}^{-3}\ \text{sec}^{-1}$
• D. $0.033\ \text{mol dm}^{-3}\ \text{sec}^{-1}$

Hint:

Look at the coefficients: 2 moles of $\text{SO}_4^{2-}$ are formed for every 1 mole of $\text{I}_3^-$. This means $\text{I}_3^-$ forms at exactly half the speed of $\text{SO}_4^{2-}$. Divide by 2 and you're done!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a balanced chemical equation and the rate of formation of one of the products ($\text{SO}^{2-}_4$). We need to determine the rate of formation of another product ($\text{I}^-_3$) using stoichiometric relationships.

Step 2: Key Formula or Approach:
From the principles of chemical kinetics, the overall rate of a reaction can be expressed in terms of the rate of change of concentrations of its components, divided by their respective stoichiometric coefficients: $$ \text{Rate} = \frac{d[\text{I}_3^-]}{dt} = \frac{1}{2}\frac{d[\text{SO}_4^{2-}]}{dt} $$

Step 3: Detailed Explanation:
The given rate of formation of $\text{SO}_4^{2-}$ is: $$ \frac{d[\text{SO}_4^{2-}]}{dt} = 0.022\ \text{mol dm}^{-3}\ \text{sec}^{-1} $$ According to our kinetic relationship: $$ \frac{d[\text{I}_3^-]}{dt} = \frac{1}{2} \times \left(\frac{d[\text{SO}_4^{2-}]}{dt}\right) $$ Substituting the given value: $$ \frac{d[\text{I}_3^-]}{dt} = \frac{1}{2} \times 0.022 = 0.011\ \text{mol dm}^{-3}\ \text{sec}^{-1} $$

Step 4: Final Answer:
The rate of formation of $\text{I}_{3\text{(aq)}}^-$ is $0.011\ \text{mol dm}^{-3}\ \text{sec}^{-1}$, which matches option (C).