JEE MAIN - Chemistry (2022 - 26th June Morning Shift - No. 15)
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
200
Explanation
$\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$
$\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$
$\mathrm{A}_0=\mathrm{B}_0$
$\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}$
$\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$
$\mathrm{t}=200 ~ \mathrm{sec}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$
$\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$
$\mathrm{A}_0=\mathrm{B}_0$
$\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}$
$\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$
$\mathrm{t}=200 ~ \mathrm{sec}$
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