A solution of H_2O_2 when titrated against KMnO_4 solution at dif… - Vidyadip

Question

A solution of $\ce{H_2O_2}$ when titrated against $\ce{KMnO_4}$ solution at different intervals of time gave the following results:

Time $($minutes$)$	$0$	$10$	$20$
Volume of $\ce{KMnO_4 (mL)}$	$23.8$	$14.7$	$9.1$

Show that decomposition of $\ce{H_2O_2}$ is first order reaction.

✓

Answer

$\text{k}=\frac{2.303}{14.7}\log\frac{2.303}{10}=\frac{2.303}{10}\times0.2093=0.048\min^{-1}$
$\text{k}=\frac{2.303}{20}\log\frac{2.303}{9.1}=\frac{2.303}{20}\times0.4176=0.048\min^{-1}$

Since the value of $k$ comes out to be constant in both the cases.
therefore the reaction is of first order.

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