Ncert Exercises & Solutions

The correct expression for the rate of reaction given below is:
$5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_2(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

1.	$\frac{\Delta\left[B r^{-}\right]}{\Delta t}=5 \frac{\Delta\left[H^{+}\right]}{\Delta t}$	2.	$\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta t}=\frac{6}{5} \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta t}$
3.	$\frac{\Delta[\mathrm{Br^-}]}{\Delta t}=\frac{5}{6} \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta t}$	4.	$\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta t}=6 \frac{\Delta\left[\mathrm{H}^{+}\right]}{\Delta t}$

HINT: Rate of reaction =

- \frac{1}{s t o i c h i o m e t r i c c o e f f i c i e n t o f r e a c t a n t} \times \frac{d [c o n c e n t r a t i o n o f r e a c t a n t]}{d T}

$- \frac{1}{stoichiometric \ coefficient\ of reactant } \times \frac{d \left[concentration \ of \ reactant\right]}{dT}$

Explanation:

The chemical reaction is as follows:

$5 B r^{-} \left(\right. a q \left.\right) + B r O_{3}^{-} \left(\right. a q \left.\right) + 6 H^{+} \left(\right. a q \left.\right) \rightarrow 3 B r_{2} \left(\right. a q \left.\right) + 3 H_{2} O \left(\right. l \left.\right)$

Rate law expression for the above equation can be written as

$\begin{aligned}- \frac{1}{5} \frac{\Delta \left[\right. B r^{-} \left]\right.}{\Delta t} &= - \frac{\Delta \left[\right. B r O_{3}^{-} \left]\right.}{\Delta t} = \frac{- 1}{6} \frac{\Delta \left[\right. H^{+} \left]\right.}{\Delta t} = \frac{+ 1}{3} \frac{\Delta \left[\right. B r_{2} \left]\right.}{\Delta t} \\ \Rightarrow \frac{\Delta \left[\right. B r^{-} \left]\right.}{\Delta t} &= - \frac{\Delta \left[\right. B r O_{3}^{-} \left]\right.}{\Delta t} = \frac{- 5}{6} \frac{\Delta \left[\right. H^{+} \left]\right.}{\Delta t} \\ \Rightarrow \frac{\Delta \left[\right. B r^{-} \left]\right.}{\Delta t} &= \frac{5}{6} \frac{\Delta \left[\right. H^{+} \left]\right.}{\Delta t}\end{aligned}$

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