Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fifty minutes the concentration of $H_{2} O_{2}$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches 0.05 M, the rate of formation of $O_{2}$ will be:

Statement-1: For the concentration of

H_{2} O_{2}

, slow evaporation is preferred.
Statement-2: Rapid evaporation of

H_{2} O_{2}

causes the decomposition of

H_{2} O_{2}

into

H_{2}

and

O_{2}

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Decomposition of H2O2 follows first order kinetics. In fifty minutes, the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:

The correct option is A 6.93×10−4 mol min−1H2O2→H2O+12O2 2t12=50 mins t12=25 mins ∴k=0.69325 −d[H2O2]dt=d[O2]dt×2 ∴d[O2]dt=12×0.69325×0.05 =6.93×10−4

Decomposition of $H_{2} O_{2}$ follows first order kinetics. In fifty minutes, the concentration of $H_{2} O_{2}$ decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $H_{2} O_{2}$ reaches 0.05 M, the rate of formation of $O_{2}$ will be: