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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)
kA=ln 2100 ; kB=ln 250\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}kA=100ln 2 ; kB=50ln 2
At=A0 × e−kA t\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}At=A0 × e−kA 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)}At=A0 × e(100−ln 2 × t)
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)}Bt=B0 × e(50−ln 2 × t)
A0=B0\mathrm{A}_0=\mathrm{B}_0A0=B0
& At=4 Bt\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}& At=4 Bt
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}}e−100ln 2 × t=4 × e−50ln 2 × t
eln 2100 × t=4\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4e100ln 2 × t=4
ln 2100 × t=ln 4=2 ln 2\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2100ln 2 × t=ln 4=2 ln 2
t=200 sec\mathrm{t}=200 ~ \mathrm{sec}t=200 sec
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