Question

For the raction :-
$4KClO_{3} \rightarrow 3KClO_{4}+KCl$
if $\frac{- d \left[KClO_{3}\right]}{dt}=K_{1}\left[KClO_{3}\right]^{n}$
$\frac{d \left[KClO_{4}\right]}{dt}=K_{2}\left[KClO_{3}\right]^{n}$
$\frac{d \left[KCl\right]}{dt}=K_{3}\left[KClO_{3}\right]^{n}$
the correct relation between $K_{1},K_{2}$ and $K_{3}$ is :-

• A. $K_{1}=K_{2}=K_{3}$
• B. $4K_{1}=3K_{2}=K_{3}$
• C. $3K_{1}=4K_{2}=12K_{3}$
• D. $12K_{1}=4K_{2}=3K_{3}$

Answer 1

Answer: (C)

$\frac{1}{4}\left(\frac{- d \left[\left(KClO\right)_{3}\right]}{dt}\right)=\frac{1}{3}\frac{d \left[\left(KClO\right)_{4}\right]}{dt}=\frac{d \left[KCl\right]}{dt}$
$\frac{K_{1}}{4}=\frac{K_{2}}{3}=\frac{K_{3}}{1}$
$3K_{1}=4K_{2}=12K_{3}$