A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is _____

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A flask is filled with equal moles of $A$ and $B$ . The half lives of $A$ and $B$ are $100 s$ and $50 s$ respectively and are independent of the initial concentration. The time required for the concentration of $A$ to be four times that of $B$ is ______ $s$ .

( Given: $\ln 2 = 0.693$ )

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answer is 200.

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Detailed Solution

$k_{A} = \frac{\ln 2}{100}; k_{B} = \frac{\ln 2}{50} A_{t} = A_{0} \times e^{- k_{A} t} A_{t} = A_{0} \times e^{(\frac{- \ln 2}{100} \times t)} B_{t} = B_{0} \times e^{(\frac{- \ln 2}{50} \times t)} A_{0} = B_{0}$

And $A_{t} = 4 B_{t}$

$e^{- \frac{\ln 2}{100} \times t} = 4 \times e^{- \frac{\ln 2}{50} \times t} e^{\frac{\ln 2}{100} \times t} = 4 \frac{\ln 2}{100} \times t = \ln 4 = 2 \ln 2 t = 200 \sec$