Question

Consider the reaction

\[ 3 \mathrm{I}_{(\mathrm{aq})}^{-}+\mathrm{S}_2 \mathrm{O}_{8(\mathrm{aq})}^{--} \longrightarrow \mathrm{I}_{3(\mathrm{aq})}^{-}+2 \mathrm{SO}_{4(\mathrm{aq})}^{2-} \]

rate of formation of \(\mathrm{SO}_4^{2-}\) at a particular time is

\[ 2.2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \]

What is the value of \(\frac{d}{d t}\left[I_3^{-}\right]\)?

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

Hint:

Divide each rate by its stoichiometric coefficient: the rate of formation of \(\text{I}_3^-\) is half that of \(\text{SO}_4^{2-}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For \(3\text{I}^-+\text{S}_2\text{O}_8^{2-}\rightarrow\text{I}_3^-+2\text{SO}_4^{2-}\), reaction rate is \(\dfrac{d[\text{I}_3^-]}{dt}=\dfrac12\dfrac{d[\text{SO}_4^{2-}]}{dt}\).

Step 2: Substitute
\(\dfrac{d[\text{I}_3^-]}{dt}=\dfrac12\times2.2\times10^{-2}\).

Step 3: Result
\(=1.1\times10^{-2}\ \mathrm{mol\,dm^{-3}\,s^{-1}}\).

Final Answer:
The rate of formation of \(\text{I}_3^-\) is \(1.1\times10^{-2}\) mol dm\(^{-3}\) s\(^{-1}\). \[ \boxed{\text{(A) }1.1\times 10^{-2}\text{ mol dm}^{-3}\text{ s}^{-1}} \]