Q83 / 200 · NEET 2009 (Screening) · Chemistry
In the reaction, \(\mathrm{BrO}_3^{-}\)(aq.) \(+5 \mathrm{Br}^{-}\)(aq.) \(+6 \mathrm{H}^{+}\)(aq.) \(\rightarrow 3 \mathrm{Br}_2(\mathrm{l})+3 \mathrm{H}_2 \mathrm{O}\) (l) .The rate of appearance of bromine $\left(\mathrm{Br}_2\right)$ is related to rate of disappearance of bromide ions as following
- Correct answerA$\frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}$
- B$\frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}$
- C$\frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=\frac{5}{3} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}$
- D$\frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}$
Solution
Key Idea Rate of appearance/disappearance
$$
= \pm \frac{1}{\begin{array}{l}
\text { stoichiometric } \\
\text { coefficient }
\end{array}} \times \frac{\text { [reactant or product] }}{\text { time taken }}
$$
For reaction,
$$
\begin{aligned}
\mathrm{BrO}_3^{-}(\mathrm{aq})+5 \mathrm{Br}^{-}(\mathrm{aq})+6 \mathrm{H}^{+} \longrightarrow & 3 \mathrm{Br}_2(1) \\
& +3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})
\end{aligned}
$$
Rate of appearance of bromine $\left(\mathrm{Br}_2\right)$
$$
=+\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}
$$
Rate of disappearance of bromide ion $\left(\mathrm{Br}^{-}\right)$
$$
=-\frac{1}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}
$$
or $\quad \frac{\mathrm{d}\left[\mathrm{Br}_2\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}$
Hence, the correct answer is option: 'A'.