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(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$.Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{k_2}{k_1} = \frac{

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$17$

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(D) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$.Given $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{k_2}{k_1} = \frac{5}{2}$.For a first-order reaction,the time $t$ to complete a fraction $x$ is $t = \frac{1}{k} \ln \frac{1}{1-x}$.For Reaction $1$,$t_1 = \frac{1}{k_1} \ln \frac{1}{1 - 2/3} = \frac{1}{k_1} \ln 3$.For Reaction $2$,$t_2 = \frac{1}{k_2} \ln \frac{1}{1 - 4/5} = \frac{1}{k_2} \ln 5$.Taking the ratio: $\frac{t_1}{t_2} = \frac{k_2}{k_1} 	imes \frac{\ln 3}{\ln 5} = \frac{5}{2} 	imes \frac{\log_{10} 3}{\log_{10} 5}$.Substituting the given values: $\frac{t_1}{t_2} = \frac{5}{2} 	imes \frac{0.477}{0.699} = 2.5 	imes 0.6824 = 1.706$.Rounding to the nearest integer,$1.706 \approx 1.7 = 17 	imes 10^{-1}$.

$S$.No.	Time	Total pressure in Pascal
$1$.	At the end of $10 \ min$	$300$
$2$.	After completion	$200$

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