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(C) The rate constants for first-order reactions are given by $k = \frac{\ln 2}{t_{1/2}}$.$k_{A} = \frac{\ln 2}{100} \, s^{-1}$ and $k_{B} = \frac{\ln

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$200$

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(C) The rate constants for first-order reactions are given by $k = \frac{\ln 2}{t_{1/2}}$.$k_{A} = \frac{\ln 2}{100} \, s^{-1}$ and $k_{B} = \frac{\ln 2}{50} \, s^{-1}$.The concentration at time $t$ is given by $[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$.Given $[A]_0 = [B]_0$,we set $[A]_t = 4[B]_t$.$[A]_0 e^{-k_A t} = 4 [A]_0 e^{-k_B t}$.$e^{(k_B - k_A)t} = 4$.Taking natural log on both sides: $(k_B - k_A)t = \ln 4 = 2 \ln 2$.$(\frac{\ln 2}{50} - \frac{\ln 2}{100})t = 2 \ln 2$.$(\frac{2 \ln 2 - \ln 2}{100})t = 2 \ln 2$.$(\frac{\ln 2}{100})t = 2 \ln 2$.$t = 200 \, s$.

$A$ 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 : $\ln 2 = 0.693$ )

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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 : $\ln 2 = 0.693$ )