What is the numerical value of the rate const | vedclass

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(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given that the reaction i

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$0.0223$

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(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given that the reaction is $20 \%$ completed,the remaining concentration $[A]_t$ is $80 \%$ of the initial concentration $[A]_0$. So,$[A]_t = 0.80 [A]_0$ and $t = 10 \ min$. Substituting these values: $k = \frac{2.303}{10} \log \frac{1}{0.8} = \frac{2.303}{10} \log(1.25)$. Since $\log(1.25) \approx 0.0969$,we get $k = \frac{2.303 	imes 0.0969}{10} \approx 0.0223 \ min^{-1}$.

What is the numerical value of the rate constant of a first-order reaction that is $20 \%$ completed in $10 \ min$ (in $min^{-1}$)?

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