Verified Solution • 2025–26 Session moderate

QJEEM22Ch57d0: JEE Main 2022 Chemistry - Class 12 JEEM22Ch57d0 Solution: A flask is filled with equal moles of A and B. The half lives…

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)

Step-by-Step Verified Solution

kA=ln⁡2100;kB=ln⁡250\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}

At=A0×e−kAt\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}

At=A0×e(−ln⁡2100×t)\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}

Bt=B0×e(−ln⁡250×t)\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}

A0=B0\mathrm{A}_0=\mathrm{B}_0

& At=4 Bt\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}

e−ln⁡2100×t=4×e−ln⁡250×t\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}

eln⁡2100×t=4\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4

eln⁡2100×t=4\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4

ln⁡2100×t=ln⁡4=2ln⁡2\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2

t=200 sec\mathrm{t}=200 ~ \mathrm{sec}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring