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(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/8}$,the amount decomposed is $x =

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$9$

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(C) For a first-order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/8}$,the amount decomposed is $x = a/8$,so $a-x = 7a/8$. Thus,$K = \frac{2.303}{t_{1/8}} \log \frac{a}{7a/8} = \frac{2.303}{t_{1/8}} \log(8/7)$.For $t_{1/10}$,the amount decomposed is $x = a/10$,so $a-x = 9a/10$. Thus,$K = \frac{2.303}{t_{1/10}} \log \frac{a}{9a/10} = \frac{2.303}{t_{1/10}} \log(10/9)$.Equating the two expressions for $K$: $\frac{1}{t_{1/8}} \log(8/7) = \frac{1}{t_{1/10}} \log(10/9)$.$\frac{t_{1/8}}{t_{1/10}} = \frac{\log(8/7)}{\log(10/9)} = \frac{3\log 2 - \log 7}{\log 10 - \log 3^2} = \frac{3(0.3) - 0.845}{1 - 2(0.477)} = \frac{0.9 - 0.845}{1 - 0.954} = \frac{0.055}{0.046} \approx 1.2$.Note: The provided solution in the prompt was mathematically incorrect. Based on standard kinetics,the ratio is approximately $1.2$,so $\frac{t_{1/8}}{t_{1/10}} 	imes 10 \approx 12$.

$t/\text{min}$	$[A]/M$
$0$	$0.6500$
$x$	$0.0650$
$20$	$0.00065$

$A$ first-order reaction decomposes such that the time taken for $1/8$ and $1/10$ of the initial concentration to decompose are $t_{1/8}$ and $t_{1/10}$ respectively. Find the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$. (Given: $\log_{10} 2 = 0.3$)

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