The rate of a first-order reaction is 0.04 m | vedclass

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(A) For a first-order reaction,the rate $r = k[A]_t = k[A]_0 e^{-kt}$.Given $r_1 = 0.04 \ mol \ L^{-1} \ s^{-1}$ at $t_1 = 10 \ min$ and $r_2 = 0.03 \

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$24$

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(A) For a first-order reaction,the rate $r = k[A]_t = k[A]_0 e^{-kt}$.Given $r_1 = 0.04 \ mol \ L^{-1} \ s^{-1}$ at $t_1 = 10 \ min$ and $r_2 = 0.03 \ mol \ L^{-1} \ s^{-1}$ at $t_2 = 20 \ min$.$\frac{r_1}{r_2} = \frac{k[A]_0 e^{-kt_1}}{k[A]_0 e^{-kt_2}} = e^{k(t_2 - t_1)}$.$\frac{0.04}{0.03} = e^{k(20 - 10)} = e^{10k}$.$\frac{4}{3} = e^{10k} \implies 10k = \ln(\frac{4}{3}) = 2.303 \log(\frac{4}{3})$.$10k = 2.303 	imes (2 \log 2 - \log 3) = 2.303 	imes (2 	imes 0.3010 - 0.4771) = 2.303 	imes (0.6020 - 0.4771) = 2.303 	imes 0.1249 \approx 0.2876$.$k = 0.02876 \ min^{-1}$.Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.02876} \approx 24.09 \ min$.Thus,the half-life is approximately $24 \ minutes$.

The rate of a first-order reaction is $0.04 \ mol \ L^{-1} \ s^{-1}$ at $10 \ minutes$ and $0.03 \ mol \ L^{-1} \ s^{-1}$ at $20 \ minutes$ after initiation. The half-life of the reaction is . . . . . . minutes. (Given $\log 2 = 0.3010, \log 3 = 0.4771$)

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