In the reaction, ${B r O_{3}}^{-} (a q .) + 5 {B r}^{-} (a q .) + 6 H^{+} (a q .) ⟶ 3 {B r}_{2} (l) + 3 H_{2} O (l)$ .
The rate of appearance of bromine $({B r}_{2})$ is related to rate of disappearance of bromide ions as following:

\frac{d [{B r}_{2}]}{d t} = \frac{2}{5} \frac{d [{B r}^{-}]}{d t}

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\frac{d [{B r}_{2}]}{d t} = - \frac{3}{5} \frac{d [{B r}^{-}]}{d t}

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\frac{d [{B r}_{2}]}{d t} = - \frac{5}{3} \frac{[{B r}^{-}]}{d t}

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\frac{d [{B r}_{2}]}{d t} = + \frac{5}{3} \frac{d [{B r}^{-}]}{d t}

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Solution

The correct option is D $\frac{d [{B r}_{2}]}{d t} = - \frac{3}{5} \frac{d [{B r}^{-}]}{d t}$
$- \frac{1}{5} \frac{d}{d t} [{B r}^{-}] = \frac{1}{3} \frac{d}{d t} [{B r}_{2}] \frac{d}{d t} [{B r}_{2}] = - \frac{3}{5} \frac{d}{d t} [{B r}^{-}]$

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Similar questions

B r O_{3}^{-} (a q) + 5 B r^{- 1} (a q) + 6 H^{+} \to 3 B r_{2} (l) + 2 H_{2} O (l)

B r O_{3 (a q)}^{-} + 5 B r_{(a q)}^{-} + 6 H_{(a q)}^{+} \to 3 B r_{2 (l)} + 3 H_{2} O_{(l)}

In the following reaction, how is the rate of appearance of the product Bromine related to the rate of disappearance of the reactant Bromide?

$B r O_{3 (a q)}^{⊝} + 5 B r_{(a q)}^{⊝} + 6 H_{(a q)}^{\oplus} \to 3 B r_{2 (l)} + 3 H_{2} O_{(l)}$

B r O_{3}^{-} (a q) + 5 B r^{-} (a q) + 6 H^{+} \to 3 B r_{2} (l) + 3 H_{2} O (l)

\frac{d [B r_{2}]}{d t} = k [B r^{-}] [B r O_{3}] [H^{+}]^{2}

- \frac{d [B r^{-}]}{d t} = k^{'} [B r^{-}] [B r O_{3}] [H^{+}]^{2}

B r_{2}

{B r}^{-}

B r O^{3 -} (a q) + 5 B r^{-} (a q) + 6 H^{+} \to 3 B r_{2} (l) + 3 H_{2} O (l)

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