Q26 / 160 · TS EAMCET 2016 · Chemistry
For the reaction
$$
\begin{gathered}
5 \mathrm{Br}^{-}(\alpha q)+6 \mathrm{H}^{+}(\alpha q)+\mathrm{BrO}_3^{-}(\alpha q) \longrightarrow \\
3 \mathrm{Br}_2(\alpha q)+3 \mathrm{H}_2 \mathrm{O}(l) \\
\text { if, }-\frac{\Delta\left[\mathrm{BrO}_3^{-}\right]}{\Delta t}=0.01 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}, \\
\frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta t} \text { in } \mathrm{mol} \mathrm{L} \\
\end{gathered}
$$
- A0.01
- B0.3
- Correct answerC0.03
- D0.005
Solution
$$
\begin{aligned}
& 5 \mathrm{Br}^{-}(a q)+6 \mathrm{H}^{+}(a q)+\mathrm{BrO}_3^{-}(a q) \longrightarrow 3 \mathrm{Br}_2(a q) \\
& +3 \mathrm{H}_2 \mathrm{O}(t) \\
& \text { Since, we've }-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}=-\frac{\Delta\left[\mathrm{Bro}_3^{-}\right]}{\Delta \mathrm{t}}=+\frac{1}{3} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}} \\
& \therefore \quad \frac{\Delta \mathrm{BrO}^{-} 3}{\Delta \mathrm{t}}=\frac{1}{3} \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}} \\
& \Rightarrow \\
& \frac{\Delta\left[\mathrm{Br}_2\right]}{\Delta \mathrm{t}}=3 \frac{\Delta\left[\mathrm{BrO}_3^{-}\right]}{\Delta \mathrm{t}}=3 \times 0.01=0.03 \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.