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(A) For a first-order reaction,the rate constant $k$ is given by the formula:$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}ight)$At $t = 20 	ext{ mi

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$10$

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(A) For a first-order reaction,the rate constant $k$ is given by the formula:$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}ight)$At $t = 20 	ext{ min}$,$[A]_0 = 0.6500 	ext{ M}$ and $[A]_t = 0.00065 	ext{ M}$.$k = \frac{1}{20} \ln\left(\frac{0.6500}{0.00065}ight) = \frac{1}{20} \ln(1000) = \frac{1}{20} 	imes 6.908 = 0.3454 	ext{ min}^{-1}$.Now,for $t = x$,$[A]_t = 0.0650 	ext{ M}$.$k = \frac{1}{x} \ln\left(\frac{[A]_0}{[A]_t}ight)$$0.3454 = \frac{1}{x} \ln\left(\frac{0.6500}{0.0650}ight)$$0.3454 = \frac{1}{x} \ln(10)$$0.3454 = \frac{2.303}{x}$$x = \frac{2.303}{0.3454} \approx 6.67 	ext{ min}$.Rounding to the nearest integer,we get $x = 7 	ext{ min}$. However,looking at the options provided,there might be a discrepancy in the question data or options. If we consider the ratio of concentrations,$[A]_0/[A]_x = 10$ and $[A]_0/[A]_{20} = 1000 = 10^3$. Since the reaction is first order,the time taken for a concentration change of $10^n$ is proportional to $n$. Thus,if $20 	ext{ min}$ corresponds to $n=3$,then $x$ (for $n=1$) should be $20/3 \approx 6.67 	ext{ min}$. Given the options,$x = 7$ is the closest integer.

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

For a first order reaction $A \rightarrow B$,the data is provided in the table below. Find $x$ in minutes. (Nearest integer)
$t/\text{min}$ $[A]/M$
$0$ $0.6500$
$x$ $0.0650$
$20$ $0.00065$

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For a first order reaction $A \rightarrow B$,the data is provided in the table below. Find $x$ in minutes. (Nearest integer)$t/\text{min}$$[A]/M$$0$$0.6500$$x$$0.0650$$20$$0.00065$