Question

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

Answer: 108

$\Rightarrow$ To calculate $:\left[A_t\right]=16\times \left[B_t\right]\dots .(1)$ time $=?$

$\to$ For I order kinetic : $\left[A_t\right]=\frac{A_0}{(2{)}^n}$

$n\to$ no of Half lives

$\Rightarrow$ Now from the relation (1)

$\left[A_t\right]=16\times \left[B_t\right]$

$\Rightarrow \frac{x}{(2{)}^{n_1}}=\frac{x}{(2{)}^{n_2}}\times 16$

$\Rightarrow (2{)}^{n_2}=(2{)}^{n_1}\times (2{)}^4$

$\Rightarrow n_2=n_1+4$

$\Rightarrow \frac{t}{{\left(t_{1/2}\right)}_2}=\frac{t}{{\left(t_{1/2}\right)}_1}+4$

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

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

$\Rightarrow t=108min$