A first order reaction takes 40 min for 20 % | vedclass

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

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$5.6 	imes 10^{-3} \ min^{-1}$

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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 the reaction is $20 \%$ decomposed,if the initial concentration $[A]_0 = 100$,then the remaining concentration $[A]_t = 100 - 20 = 80$.Substituting the values into the formula: $k = \frac{2.303}{40 \ min} \log_{10} \frac{100}{80}$.$k = \frac{2.303}{40} \log_{10} (1.25)$.Using $\log_{10} (1.25) \approx 0.0969$,we get $k = \frac{2.303 	imes 0.0969}{40} \ min^{-1}$.$k \approx 0.00557 \ min^{-1} = 5.57 	imes 10^{-3} \ min^{-1}$.Rounding to two significant figures,$k \approx 5.6 	imes 10^{-3} \ min^{-1}$.

$A$ first order reaction takes $40 \ min$ for $20 \%$ decomposition. Calculate its rate constant.

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