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

The time taken for the concentration of A to become $16$ times that of B is $\mathbf{108}\mathbf{}\mathbf{min}\mathbf{.}$

Explanation:

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Given}\mathbf{}\mathbf{values} \\ \mathrm{A}\stackrel{\mathrm{first}\mathrm{order}}{\to }{\mathrm{t}}_{\frac{1}{2}}\left(\mathrm{A}\right)=54 \\ \mathrm{B}\stackrel{\mathrm{first}\mathrm{order}}{\to }\mathrm{t}\frac{1}{2}\left(\mathrm{B}\right)=18 \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{time}\mathbf{} \\ {\mathrm{A}}_0={\mathrm{B}}_0={\mathrm{C}}_0 \\ {\mathrm{A}}_{\mathrm{t}}=\frac{{\mathrm{A}}_0}{\frac{2\mathrm{t}}{54}} \\ {\mathrm{B}}_{\mathrm{t}}=\frac{{\mathrm{B}}_0}{\frac{2\mathrm{t}}{18}} \\ \Rightarrow {\mathrm{A}}_{\mathrm{t}}=16\times {\mathrm{B}}_{\mathrm{t}} \\ \Rightarrow \frac{{\mathrm{A}}_0}{\frac{2\mathrm{t}}{54}}=16\times \frac{{\mathrm{B}}_0}{\frac{2\mathrm{t}}{18}} \\ \Rightarrow 2^{\mathrm{t}/18-\mathrm{t}/54} \\ \Rightarrow 2^{2\mathrm{t}/54}=16=2^4 \\ \Rightarrow 2\mathrm{t}/54=4 \\ \Rightarrow \mathrm{t}=\mathbf{108}\mathbf{}\mathbf{min} \\ \mathrm{Hence},\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{is}\mathbf{108}\mathbf{}\mathbf{min}\mathbf{.} \end{gathered}$$