Chemical Kinetics Question 13
Question 13 - 01 February - Shift 1
$A$ and $B$ are two substances undergoing radioactive decay in a container. The half life of $A$ is $15 min$ and that of $B$ is $5 min$. If the initial concentration of $B$ is 4 times that of $A$ and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? $\min$.
Show Answer
Answer: (15)
Solution:
$[A] _{t}=[A]_0 e^{-kt}$
For $\mathbf{A}$ : Let $[A] _{t}$ be $y$ and $[A]_0$ be $x ; k=\frac{\ln 2}{t _{1 / 2}}=$
$\ln 2$
$15 min$
$y=x e^{-k t}$
$=x e^{-(\frac{\ln 2}{15}) t}$
For $\mathbf{B}:[B] _{t}=[B]_0 e^{-kt}$
Let $[B] _{t}=y ;[B]_0=4 x ; k=\frac{\ln 2}{t _{1 / 2}}=\frac{\ln 2}{5 \min }$
$y=4 x e^{-(\frac{\ln 2}{5}) t}$
$x e^{-(\frac{\ln 2}{15}) t} \quad 4 x e^{-(\frac{\ln 2}{5}) t}$
$e^{t(\frac{\ln 2}{5}-\frac{\ln 2}{15})}=4$
$t \times[\frac{\ln 2}{5}-\frac{\ln 2}{15}]=\ln 4$
$t \times \ln 2[\frac{1}{5}-\frac{1}{15}]=2 \ln 2$
$t=15 min$
