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(C) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: Initial concentration $[A]_0 =

Vedclass · Accepted Answer

$4.606 	imes 10^{-3} \ s^{-1}$

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(C) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: Initial concentration $[A]_0 = \frac{M}{10} = 0.1 \ M$,Final concentration $[A]_t = \frac{M}{100} = 0.01 \ M$,and time $t = 8 \ 	ext{minutes} = 8 	imes 60 \ s = 480 \ s$. Substituting the values: $K = \frac{2.303}{480} \log \frac{0.1}{0.01} = \frac{2.303}{480} \log 10$. Since $\log 10 = 1$,$K = \frac{2.303}{480} \approx 4.8 	imes 10^{-3} \ s^{-1}$. Note: Based on the provided options,the closest match is $4.606 	imes 10^{-3} \ s^{-1}$ (Option $C$),which assumes $t = 500 \ s$ instead of $480 \ s$.

$A$ first-order reaction starts with a decimolar solution of the reactant. After $8$ minutes,its concentration becomes $M/100$. Calculate the rate constant of the reaction.

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