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 M to 0.125 Min one such decomposition. When the concentration of $H_{2} O_{2}$ reached 0.05 M, the rate of formation of $O_{2}$ will be

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$2.66 L \min^{- 1} at STP$

$6.93 \times 10^{- 4} {molL}^{- 1} \min^{- 1}$

$6.93 \times 10^{- 2} {molI}^{- 1} \min^{- 1}$

$1.34 \times 10^{- 2} mol L^{- 1} \min^{- 1}$

answer is A.

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Detailed Solution

The decomposition reaction is $H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$

The decreasing concentration from 0.5 M to 0.125 M implies two half-lives:

$0.5 M \overset{}{\overset{t_{1 / 2}}{\to}} 0.25 M \overset{}{\overset{t_{1 / 2}}{\to}} 0.125 M$

Hence, $2 t_{1 / 2} = 50 \min$ i.e. $t_{1 / 2} = 25 \min$

The rate constant of the reaction is

$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{25 \min} 2.772 \times 10^{- 2} \min^{- 1}$

The rate expression is $\frac{1}{(1 / 2)} \frac{d [O_{2}]}{d t} = k [H_{2} O_{2}]$

When $[H_{2} O_{2}] = 0.05 M$ , the rate of formation of $O_{2}$ will be

$\frac{d [O_{2}]}{d t} = \frac{1}{2} (2.772 \times 10^{- 2} min^{- 1}) (0.05 {molL}^{- 1}) = 6.93 \times 10^{- 4} mol L^{- 1} \min^{- 1}$ .

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