Question

The decomposition of hydrogen peroxide in an aqueous solution is a first order reaction, it can be studied by titrating quickly $10mL$ portions of reactions mixture at various times from the $t=0$ of reaction against a standard solution of $KMn{O}_4$. Volume of $KMn{O}_4$ solutions used in each case is proportional to the remaining concentration of $H_2{O}_2$
From the following data calculate the rate constant of the reaction

Time (seconds)	0	600	1200
$KMn{O}_4$ solution used (mL)	22.8	13.8	8.2

Answer 1

Given that decomposition of $H_2{O}_2$ in aqueous solution follows first order kinetics.

For first order reactions,

$K=\frac{1}{t}ln\frac{[A{]}_0}{[A{]}_t}$

where, $K$ is the rate constant,

$t$= time

$[A{]}_0$= Initial concentration of reactant (i.e. $H_2{O}_2$)

$[A{]}_t$= concentration of reactant at $t=t$

Now, concentration of $H_2{O}_2$ present at any time is proportional to volume of $KMn{O}_4$ used in titration.

Now, given that

at, $t=0$; volume of $KMn{O}_4$ used=$22.8mL\propto \left[A\right]$

At $t=600s$;Volume of $KMn{O}_4$ used=$13.8mL\propto [A{]}_t$

So, $K=\frac{1}{600}ln\left(\frac{22.8}{13.8}\right)$

$\Rightarrow K=\frac{1}{600}\times 0.5=0.00084s^{-1}$