In an experiment analyzing the catalytic decay of an aqueous uniform mixture of , fixed aliquots are — Chemical Kinetics Chemistry Question
Question
In an experiment analyzing the catalytic decay of an aqueous uniform mixture of $H_2O_2$, fixed $10\text{ mL}$ aliquots are routinely pipetted and titrated thoroughly against standardized $KMnO_4$. At $t=0$, the required $KMnO_4$ neutralizer volume accurately equals $25.0 \text{ mL}$. Over precisely $20 \text{ minutes}$, the requisite titration volume strictly decays to exactly $12.5 \text{ mL}$. Extrapolate logically the identical titration volume mathematically required (in mL) at the $40 \text{ minute}$ boundary.
💡 Solution & Explanation
For direct $H_2O_2$ decomposition, titrating with $KMnO_4$ relies on the core principle that neutralizing volume corresponds seamlessly to the residual untempered $H_2O_2$. As the volume falls smoothly from 25.0 mL precisely down to 12.5 mL, representing identical 50% fraction depletion, it confirms the fixed theoretical half-life $t_{1/2}$ sits at exactly 20 minutes. Therefore, at cumulative elapsed timestamp $t = 40 \text{ minutes}$ (constituting 2 entire unbroken half-lives), the remaining concentration will divide in half once more: $12.5 / 2 = 6.25 \text{ mL}$.