MCQ
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 $300$ $s$ and $180\, s ,$ 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
    $300$
  • $900$

Answer

Correct option: D.
$900$
d
$[ A ]_{ t }=4[ B ]_{ t }$

$[ A ]_{0} e ^{-\left( in ^{2} / 300\right)^{t}}=4[ B ]_{0} e ^{(-\ln 2 / 180) t }$

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

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

$\left(\frac{1}{180}-\frac{1}{300}\right) t=2 \Rightarrow t=\frac{2 \times 180 \times 300}{120}=900 sec$

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