For a second-order reaction,the rate constant | vedclass

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(C) For a second-order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} \right)$.Given: $k

Vedclass · Accepted Answer

$1.25 	imes 10^4 \, 	ext{min}$

Vedclass · Answer

(C) For a second-order reaction,the integrated rate equation is given by: $k = \frac{1}{t} \left( \frac{1}{[A]_t} - \frac{1}{[A]_0} ight)$.Given: $k = 8 	imes 10^{-5} \, M^{-1} \, 	ext{min}^{-1}$,$[A]_0 = 1 \, M$,and $[A]_t = 0.5 \, M$.Substituting the values: $8 	imes 10^{-5} = \frac{1}{t} \left( \frac{1}{0.5} - \frac{1}{1} ight)$.$8 	imes 10^{-5} = \frac{1}{t} (2 - 1)$.$8 	imes 10^{-5} = \frac{1}{t} 	imes 1$.$t = \frac{1}{8 	imes 10^{-5}} = \frac{10^5}{8} = 0.125 	imes 10^5 = 1.25 	imes 10^4 \, 	ext{min}$.

For a second-order reaction,the rate constant is $8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$. In how much time will a $1 \, M$ solution decrease to $0.5 \, M$?

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