Question

A and $B$ decompose via first order kinetics with half-lives $54.0\,min$ and $18.0$ 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

$\Rightarrow $ To calculate $:\left[ A _{ t }\right]=16 \times\left[ B _{ t }\right] \ldots .(1)$ time $=?$

$ \rightarrow$ For I order kinetic : $\left[ A _{ t }\right]=\frac{ A _{0}}{(2)^{ n }}$

$n \rightarrow$ no of Half lives

$\Rightarrow $ Now from the relation (1)

$\left[ A _{ t }\right]=16 \times\left[ B _{ t }\right]$

$\Rightarrow \frac{ x }{(2)^{ n _{1}}}=\frac{ x }{(2)^{ n _{2}}} \times 16$

$ \Rightarrow (2)^{ n _{2}}=(2)^{ n _{1}} \times(2)^{4}$

$\Rightarrow n_{2}=n_{1}+4 $

$\Rightarrow \frac{t}{\left(t_{1 / 2}\right)_{2}}=\frac{t}{\left(t_{1 / 2}\right)_{1}}+4$

$\Rightarrow t \left(\frac{1}{18}-\frac{1}{54}\right)=4$

$ \Rightarrow t =\frac{4 \times 18 \times 54}{36}$

$\Rightarrow t =108\,\,min$