Two substances A $(t_{1 / 2} = 5 \min)$ and B $(t_{1 / 2} = 15 \min)$ are taken in such a way that initially [A]= 4[B]. Th time after which both the concentrations will be equal is : (Assume that reaction is first order)

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5 min

20 min

Concentration can never he equal

15 min

answer is B.

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Detailed Solution

For the given substance,

$[A]_{t} = [B]_{t}$

$\begin{aligned} [A]_{0} e^{- k_{1} t} & = [B]_{0} e^{- k_{2} t} \\ 4 & = e^{(k_{1} - k_{2}) t} \\ t & = \frac{2 \ln 2}{\frac{\ln 2}{5} - \frac{\ln 2}{15}} = 15 minutes \end{aligned}$

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