For the second order reaction, if the concentration of reactant changes from $0.08 M$ to $0.04 M$ in 10 minutes. Calculate the time at which concentration of reactant becomes $0.01 M$ .

20 m i n

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30 m i n

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50 m i n

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70 m i n

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Solution

The correct option is D $70 m i n$
For a second order reaction we have
$\frac{1}{{[A]}_{t}} - \frac{1}{[A]_{o}} = k t$
In $10$ minutes, concentration changes from $0.08 M$ to $0.04 M$
$(\frac{1}{0.04} - \frac{1}{0.08}) \frac{1}{10} = k {m i n}^{- 1} 1.25 {m i n}^{- 1} = k$
Substituting value of $k$ , we get it become $(\frac{1}{0.01} - \frac{1}{0.08}) \frac{1}{1.25} = t 70 m i n = t$
Hence answer is $[D]$

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For the second order reaction, if the concentration of reactant changes from 0.08M to 0.04M in 10 minutes. Calculate the time at which concentration of reactant becomes 0.01M.

The correct option is D 70minFor a second order reaction we have 1[A]t−1[A]o=kt In 10 minutes, concentration changes from 0.08M to 0.04M(10.04−10.08)110=kmin−11.25min−1=k Substituting value of k, we get it become (10.01−10.08)11.25=t70min=t Hence answer is [D]

For the second order reaction, if the concentration of reactant changes from $0.08 M$ to $0.04 M$ in 10 minutes. Calculate the time at which concentration of reactant becomes $0.01 M$ .