Question

A flask is filled with equal moles of $A$ and $B$. The half lives of $A$ and $B$ are $100s$ and $50s$ 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

Answer: 200

$k_A=\frac{\ln 2}{100};k_B=\frac{\ln 2}{50}$
$A_t=A_0\times e^{-k_{A}t}$
$A_t=A_0\times e^{(\frac{-\ln 2}{100}\times t})$
$B_t=B_0\times e^{\left(\frac{-\ln 2}{50}\times t\right)}$
$A_0=B_0$
$\&A_t=4B_t$
$e^{-\frac{\ln 2}{100}\times t}=4\times e^{-\frac{\ln 2}{50}\times t}$
$e^{\frac{\ln 2}{100}\times t}=4$
$e^{\frac{\ln 2}{100}\times t}=4$
$\frac{\ln 2}{100}\times t=\ln 4=2\ln 2$
$t=200\sec$