Half-life is independent of the concentration of the reactant. After $10$ minutes, the volume of $N_{2}$ gas is $10 L$ and after complete reaction, it is $50 L$ . Hence, the rate constant is:

(2.303 / 10) l o g 5

m i n^{- 1}

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(2.303 / 10) l o g 1.25 m i n^{- 1}

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(2.303 / 10) l o g 2 m i n^{- 1}

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(2.303 / 10) l o g 4 m i n^{- 1}

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Solution

The correct option is B $(2.303 / 10) l o g 1.25 m i n^{- 1}$

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Half - life of the reaction : $H_{2} O_{2 (a q)} \to H_{2} O_{ℓ} + \frac{1}{2} O_{2 (g)}$ is independent of initial concentration of $H_{2} O_{2}$ . Volume of $O_{2}$ gas after 20 minutes is 5L at 1 atm and $27^{\circ} C$ , and after completion of the reaction 50 L. The rate constant is :