A first order reaction takes 23.03 minutes f | vedclass

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(D) 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,$[A]_0 = 100 \

Vedclass · Accepted Answer

$9.69 	imes 10^{-3} \ minute^{-1}$

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(D) 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,$[A]_0 = 100 \%$,$[A]_t = 100 - 20 = 80 \%$,and $t = 23.03 \ minutes$.Substituting these values into the equation:$k = \frac{2.303}{23.03} \log_{10} \left( \frac{100}{80} ight)$.$k = 0.1 	imes \log_{10}(1.25)$.Since $\log_{10}(1.25) \approx 0.0969$,we get:$k = 0.1 	imes 0.0969 = 0.00969 \ minute^{-1}$.Therefore,$k = 9.69 	imes 10^{-3} \ minute^{-1}$.

$A$ first order reaction takes $23.03 \ minutes$ for $20 \%$ decomposition. Calculate its rate constant.

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