Decomposition of $H_{2} O_{2}$ follows a first order reaction. In fify 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:

6.93 \times 10^{- 4} m o l m i n - 1

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2.66 L m i n^{- 1} a t S T P

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1.34 \times 10^{- 2} m o l m i n^{- 1}

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6.93 \times 10^{- 2} m o l m i n^{- 1}

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Solution

The correct option is A $6.93 \times 10^{- 4} m o l m i n - 1$
Decomposition of $H_{2} O_{2}$ follows a first order reaction asf follows:
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$
Given -In $50$ minutes, the concentration is reduced to one fourth, hence $t_{1 / 2} = 25 m i n u t e s$
we know that: $k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{25}$
Rate of decomposition of $H_{2} O_{2} = k [H_{2} O_{2}] = 0.05 \times \frac{0.693}{25} = 1.39 \times 10^{- 3} m o l / m i n$
Rate of formation of oxygen is one half the rate of decomposition of $H_{2} O_{2}$
It is $\frac{1}{2} \times 1.39 \times 10^{- 3} = 6.93 \times 10^{- 4} m o l / m i n$
Option A is correct.

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Similar questions

Q. 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:

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: