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(A) For a $1^{st}$ order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$Given $t = 570 \

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$2$

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(A) For a $1^{st}$ order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} ight)$Given $t = 570 \ s$ and $[A]_t = 32 \%$ of $[A]_0$,so $\frac{[A]_0}{[A]_t} = \frac{100}{32} = 3.125$.$K = \frac{2.303}{570} \log_{10} (3.125)$Since $\log_{10} (3.125) = \log_{10} (100/32) = \log_{10} 100 - \log_{10} 32 = 2 - 5 \log_{10} 2 = 2 - 5(0.301) = 2 - 1.505 = 0.495$.$K = \frac{2.303 	imes 0.495}{570} \approx \frac{1.14}{570} = 0.002 \ s^{-1} = 2 	imes 10^{-3} \ s^{-1}$.Rounding to the nearest integer,the value is $2$.

For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$

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For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$