A first order reaction is 25 % completed in 4 | vedclass

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(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}}$.Given that the r

Vedclass · Accepted Answer

$\frac{2.303}{40} 	imes \log \frac{4}{3}$

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(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]_{t}}$.Given that the reaction is $25 \%$ completed,if the initial concentration $[A]_{0} = 100$,then the remaining concentration $[A]_{t} = 100 - 25 = 75$.The time taken $t = 40 \ min$.Substituting these values into the formula:$k = \frac{2.303}{40} \log_{10} \frac{100}{75}$.Simplifying the fraction $\frac{100}{75}$ gives $\frac{4}{3}$.Therefore,$k = \frac{2.303}{40} \log_{10} \frac{4}{3}$.

$A$ first order reaction is $25 \%$ completed in $40 \ min$. What is the rate constant $k$ for the reaction?

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