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.</p

k_A= 2100 ; k_B= 250 A_t=A_0 e^-k_A t A_t=A_0 e^(- 2100 t) B_t=B_0 e^(- 250 t) A_0=B_0 \& ~A_t=4 ~B_t e^- 2100 t=4 e^- 250 t e^ 2100 t=4 <p

A flask is filled with equal moles of A and B The half lives: (Chemistry)

$\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}$

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$

$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$

$\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$

$\mathrm{A}_0=\mathrm{B}_0$

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

$\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$

$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$

$\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$

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