Question

The decomposition of hydrogen peroxide in an aqueous solution is a reaction of first order. It can be followed by titrating 10 mL portions of reaction mixture at various times from the beginning of reaction against a standard solution of $KMn{O}_4$. Volume of $KMn{O}_4$ solution 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.
$\begin{array}{cccc}\text{Time (in seconds)}& 0& 600& 1200\\ KMn{O}_4\text{sol. used (in mL)}& 22.8& 13.8& 8.20\end{array}$

• A. $k=8.44\times {10}^{-4}{\text{sec}}^{-1}$
• B. $k=8.44\times {10}^{-2}{\text{sec}}^{-1}$
• C. $k=2.44\times {10}^{-6}{\text{sec}}^{-1}$
• D. $k=2.22\times {10}^{-4}{\text{sec}}^{-1}$

Answer 1

The correct option is A $k=8.44\times {10}^{-4}{\text{sec}}^{-1}$

Given that decomposition of $H_2{O}_2$ in an aqueous solution follows first order kinetics.
For first order reactions,
$k=\frac{1}{t}ln\left[A{]}_0/\right[A{]}_t$
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
$22.8mL\propto [A{]}_0$
At $t=1200s$ Volume of $KMn{O}_4$ used$8.20mL\propto [A{]}_t$
So,
$k=1/1200\times ln\left(22.8/8.2\right)\approx 1/1200=8.33\times {10}^{-4}{\text{sec}}^{-1}$
This approximation works given the options in the question, the nearest one is $8.44\times {10}^{-4}{\text{sec}}^{-1}$