What is the rate constant of a first-order re | vedclass

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(C) 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 $60 \%$ o

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$0.020 \ minute^{-1}$

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(C) 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 $60 \%$ of the reactant decomposes,the remaining concentration $[A]_t$ is $100 - 60 = 40$ if the initial concentration $[A]_0 = 100$.Substituting the values: $k = \frac{2.303}{45} \log_{10} \frac{100}{40}$.$k = \frac{2.303}{45} \log_{10} (2.5)$.Since $\log_{10} (2.5) \approx 0.3979$,we get $k = \frac{2.303 	imes 0.3979}{45}$.$k \approx 0.020 \ minute^{-1}$.

What is the rate constant of a first-order reaction if $60 \%$ of the reactant decomposes in $45 \ minutes$?

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