For the reaction 2A + B to text{product},the | vedclass

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(C) The given rate law $\frac{-d[A]}{dt} = k[A]$ indicates a first-order reaction with respect to $A$.For a first-order reaction,the integrated rate e

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$\frac{[A]_0}{e}$

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(C) The given rate law $\frac{-d[A]}{dt} = k[A]$ indicates a first-order reaction with respect to $A$.For a first-order reaction,the integrated rate equation is $[A] = [A]_0 e^{-kt}$.The half-life $t_{1/2}$ for a first-order reaction is given by $t_{1/2} = \frac{\ln 2}{k}$,which implies $k = \frac{\ln 2}{t_{1/2}}$.Substituting the given time $t = \frac{t_{1/2}}{\ln 2}$ into the rate equation:$[A] = [A]_0 e^{-k 	imes (\frac{t_{1/2}}{\ln 2})}$$[A] = [A]_0 e^{-(\frac{\ln 2}{t_{1/2}}) 	imes (\frac{t_{1/2}}{\ln 2})}$$[A] = [A]_0 e^{-1} = \frac{[A]_0}{e}$.

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

For the reaction $2A + B \to \text{product}$,the rate law is given by $\frac{-d[A]}{dt} = k[A]$. At a time $t = \frac{t_{1/2}}{\ln 2}$,what is the concentration of the reactant $A$?

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