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(A) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.Given that $80 \%$ o

Vedclass · Accepted Answer

$0.11 \ minute^{-1}$

Vedclass · Answer

(A) 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 $80 \%$ of the reactant has reacted,if the initial concentration $[A]_0 = 100$,then the remaining concentration $[A]_t = 100 - 80 = 20$.The time taken $t = 15 \ minute$.Substituting these values into the formula:$k = \frac{2.303}{15} \log_{10} \frac{100}{20}$$k = \frac{2.303}{15} \log_{10}(5)$Using $\log_{10}(5) \approx 0.699$:$k = \frac{2.303 	imes 0.699}{15}$$k \approx 0.1073 \ minute^{-1}$Rounding to two decimal places,$k \approx 0.11 \ minute^{-1}$.

Calculate the rate constant of the first order reaction if $80 \%$ of the reactant reacted in $15 \ minute$.

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