In a reaction A + B rightarrow C,initial conc | vedclass

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(D) Given: $[A]_0 = 8[B]_0$. Half-lives: $(t_{1/2})_A = 10 \ min$,$(t_{1/2})_B = 40 \ min$. Rate constants: $k_A = \frac{\ln 2}{10}$,$k_B = \frac{\ln

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$40$

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(D) Given: $[A]_0 = 8[B]_0$. Half-lives: $(t_{1/2})_A = 10 \ min$,$(t_{1/2})_B = 40 \ min$. Rate constants: $k_A = \frac{\ln 2}{10}$,$k_B = \frac{\ln 2}{40}$. For first-order kinetics,$[A]_t = [A]_0 e^{-k_A t}$ and $[B]_t = [B]_0 e^{-k_B t}$. Setting $[A]_t = [B]_t$: $[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} \implies \frac{[A]_0}{[B]_0} = e^{(k_A - k_B)t}$. Substituting values: $8 = e^{(\frac{\ln 2}{10} - \frac{\ln 2}{40})t}$. Taking natural log: $\ln 8 = (\frac{4\ln 2 - \ln 2}{40})t$. $3 \ln 2 = (\frac{3 \ln 2}{40})t$. $t = 40 \ min$.

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