Question

A flask is filled with equal moles of $A$ and $B$. The half lives of $A$ and $B$ are $100\, s$ and $50\, s$ respectively and are independent of the initial concentration. The time required for the concentration of $A$ to be four times that of $B$ is _______$s$.
(Given : $\ln 2=0.693$ )

Answer 1

$k _{ A }=\frac{\ln 2}{100} ; k _{ B }=\frac{\ln 2}{50}$
$A _{ t }= A _{0} \times e ^{- k _{ A } t }$
$\left.A _{ t }= A _{0} \times e ^{\left(\frac{-\ln 2}{100} \times t \right.}\right)$
$B _{ t }= B _{0} \times e ^{\left(\frac{-\ln 2}{50} \times t \right)}$
$A _{0}= B _{0}$
$\& A _{ t }=4 B _{ t }$
$e ^{-\frac{\ln 2}{100} \times t }=4 \times e ^{-\frac{\ln 2}{50} \times t }$
$e ^{\frac{\ln 2}{100} \times t }=4$
$e ^{\frac{\ln 2}{100} \times t }=4$
$\frac{\ln 2}{100} \times t =\ln 4=2 \ln 2$
$t =200 \,\sec$