A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolarnon reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _____ min. (Round off to the nearest Integer)

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answer is 108.

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Detailed Solution

$A \overset{1 stOrder}{\to} t_{\frac{1}{2}} (A) = 54$
$B \overset{1 s t O r d e r}{\to} t_{\frac{1}{2}} (B) = 18$
$A_{0} = B_{0} = N_{0}$
$A_{t} = \frac{A_{o}}{\frac{2 t}{4 t}}$
$B_{t} = \frac{B_{o}}{\frac{28}{18}}$
$\Rightarrow A_{t} = 16 \cdot B_{t}$
$\Rightarrow$
$\frac{A_{o}}{\frac{2 t}{54}} = 16 \times \frac{B_{0}}{\frac{28}{18}}$
$\Rightarrow 2^{t / 18 - t / 54} = 16$
$\Rightarrow 2^{2 t / 54} = 16 = 2^{4}$
$\Rightarrow 2 t / 54 = 4$
$\Rightarrow t = 108 \min$