Q.Explain the experimental determination of rate constant for the decomposition of $H_2O_2$ in aqueous solution.

The first-order rate constant for H$_2$O$_2$ decomposition is obtained by periodically titrating the remaining, un-decomposed peroxide against standard potassium permanganate and substituting the titre volumes (proportional to concentration) into the first-order rate equation.

Reaction: $2H_2O_2 \rightarrow 2H_2O + O_2$ This decomposition (catalysed, e.g., by $I^-$ or colloidal Pt in the classroom experiment) follows first-order kinetics with respect to $H_2O_2$.

Principle: Undecomposed $H_2O_2$ is a reducing agent and is oxidised by acidified $KMnO_4$ (which is reduced from purple $Mn^{7+}$ to colourless $Mn^{2+}$):

$5H_2O_2 + 2KMnO_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$

Since $KMnO_4$ reacts only with the $H_2O_2$ still present (not with the products), the volume of standard $KMnO_4$ solution required to titrate an aliquot is directly proportional to the concentration of undecomposed $H_2O_2$ at that instant.

Procedure:

1. A known volume of $H_2O_2$ solution is taken (with a trace of catalyst, e.g. KI, to initiate decomposition) at time $t=0$; a fixed-volume aliquot is immediately withdrawn, quenched (chilled/diluted to slow further reaction), and titrated against standard $KMnO_4$ to get the titre volume $V_0$ (proportional to the initial concentration $a$).
2. As the reaction proceeds, aliquots of the same fixed volume are withdrawn at successive, recorded time intervals $t_1, t_2, t_3, \ldots$ and each is similarly titrated against the same standard $KMnO_4$ to obtain titre volumes $V_1, V_2, V_3, \ldots$ (each proportional to the H$_2$O$_2$ concentration remaining, i.e. $(a-x)$, at that time). …