Question

$A$ and $B$ decompose via first order kinetics with half-lives $54.0\text{min}$ and $18.0\text{min}$ respectively. Starting from an equimolar non-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).

Answer 1

$A\stackrel{1^{st}\text{order}}{\to }{t}_{1/2}\left(A\right)=54minB\stackrel{1^{st}\text{order}}{\to }{t}_{1/2}\left(B\right)=18min$

$A_0=B_0=N_0$
$A_t=\frac{A_0}{2^{\frac{t}{54}}}$

$B_t=\frac{B_0}{2^{\frac{t}{18}}}$

according to question,
$\Rightarrow A_t=16.Bt$

$\Rightarrow \frac{A_0}{2^{\frac{t}{54}}}=16\times \frac{B_0}{2^{\frac{t}{18}}}$

$\Rightarrow 2^{t/18-t/54}=16$

$\Rightarrow 2^{2t/54}=16=2^4$
$\Rightarrow \frac{2t}{54}=4$
$\Rightarrow t=108\text{min}$