Question

$AandB$ decompose via first order kinetics with half-lives $54.0minand18.0min$ respectively. Starting from an equimolar non reactive mixture of $AandB$, the time taken for the concentration of $A$ to become $16times$ that of $B$ is _________ min.
(Round off to the Nearest Integer) Take $\ln 2=0.693$.

• A. 108
• B. 108.00
• C. 108.0

Answer 1

Answer: (A), (B), (C)

$Initially:\left[A_o\right]=\left[B_o\right]=a$

$\text{After time 't' min, [A]}=16\left[B\right]$

$\left[A\right]=\left[A_o\right]e^{-k_{A}t}$

$\left[B\right]=\left[B_o\right]e^{-k_{B}t}$

$\Rightarrow a.e^{-k_{A}t}=16a{e}^{-k_{B}t}$

$\Rightarrow e^{-(k_A-k_B)t}=16$

$\Rightarrow (k_B-k_A)t=ln16$
$\Rightarrow (k_B-k_A)t=ln{2}^4$
$\Rightarrow ln2(\frac{1}{18}-\frac{1}{54})t=4ln2(ln2=0.693)$

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

$\Rightarrow t=\frac{54\times 18\times 4}{36}=108min$