Question

From the following data estimate the order for decomposition of an aqueous solution of hydrogen peroxide:

Time (minutes)	$0$	$10$	$20$	$30$
$V\left(ml\right)$	$46.1$	$29.8$	$19.6$	$12.3$

where $V$ is the volume of potassium permanganate solution in $mL$ required to decompose a definite volume of the peroxide solution.

• A. $0$
• B. $1$
• C. $2$
• D. $1/2$

Answer 1

The correct option is B $1$

Time (min)	$0$	$10$	$20$	$30$
V (ml)	$46.1$	$29.8$	$19.6$	$12.3$

The volume of $KMn{O}_4$ corresponds to the undecomposed $H_2{O}_2$.

Hence, at $t=0$, $a=46.1ml$

at $t=t$, volume corresponds to $a-x$

Hence, $K=\frac{2.303}{t}\log \frac{a}{a-x}$

$t=10,K=\frac{2.303}{10}\log \left(\frac{46.1}{29.8}\right)=0.043$

$t=20,K=\frac{2.303}{20}\log \left(\frac{46.1}{19.6}\right)=0.0427$

$t=30,K=\frac{2.303}{30}\log \left(\frac{46.1}{12.3}\right)=0.044$

The constancy of $K$ shows that the reaction is a first order.