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For the first order reaction : $H_{2} O_{2} (a q) ⟶ H_{2} O (l) + \frac{1}{2} O_{2} (g)$
the half life of reaction is 30 min. If the volume of $O_{2}$ (g) collected at a certain pressure and temperature is 100 ml after a long time from the start of reaction, then what was the volume of $O_{2}$ (g) collected at the same pressure and temperature after 60 min from the start of reaction?

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25 ml

50 ml

75 ml

12.5 ml

answer is B.

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Detailed Solution

First order reaction given in the problem is,

$H_{2} O_{2} (a q) ⟶ H_{2} O (l) + \frac{1}{2} O_{2} (g)$

Here,

$\begin{array}{l} k \times t = \ln (\frac{V_{\infty}}{V_{\infty} - V_{t}}) \\ \frac{\ln 2}{30} \times 60 = \ln [\frac{100}{100 - V_{t}}] \\ \Rightarrow V_{t} = 75 mL \end{array}$

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