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A flask contains a mixture of compounds A and B. Both compounds decompose by fist order kinetics. The half-lives for A and B are $300 s$ and $180 s$ , respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in S) is. (Use In $2 = 0.693$ )

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180

120

300

900

answer is C.

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Detailed Solution

$A_{k} = A_{0} e^{- k_{1} t} 4 x = A_{0} e^{- k_{1} t} A_{0} = B_{0} x = 80 \times e^{- k_{2} t} 4 = e^{- k_{1} + k_{2} t} \ln 4 = (k_{2} - k_{1}) t t = \frac{lny}{(k_{2} - k_{1})} = \frac{2 \times 0.683}{0.693 (\frac{1}{180} - \frac{1}{300})}$

.t =900

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