Question

A flask contains a mixture of compounds $A$ and B. Both compounds decompose by first-order kinetics. The half-lives for $A$ and $B$ are $300s$ and $180s$, respectively. If the concentrations of A and $B$ are equal initially, the time required for the concentration of A to be four times that of B ( in s ): (Use ln $2=0.693)$

• A. 180
• B. 120
• C. 300
• D. 900

Answer 1

Answer: (D)

$[A{]}_t=4[B{]}_t$

$[A{]}_0{e}^{-{\left({\ln }^2/300\right)}^4}=4[B{]}_0{e}^{(-\ln 2/180)t}$

$e^{\left(\frac{{\ln }^2}{180}-\frac{{\ln }^2}{300}\right)}=4$

$\left(\frac{{\ln }^2}{180}-\frac{{\ln }^2}{300}\right)t=\ln 4$

$\left(\frac{1}{180}-\frac{1}{300}\right)t=2$

$\Rightarrow t=\frac{2\times 180\times 300}{120}=900sec$