2A xrightarrow{k} B is a zero-order reaction, | vedclass

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(B) For a zero-order reaction, the rate law is given by $[A]_t = [A]_0 - kt'$.Note that the stoichiometry of the reaction is $2A \rightarrow B$. The r

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(B) For a zero-order reaction, the rate law is given by $[A]_t = [A]_0 - kt'$.Note that the stoichiometry of the reaction is $2A ightarrow B$. The rate of reaction is defined as $-\frac{1}{2} \frac{d[A]}{dt} = k$.Thus, $-\frac{d[A]}{dt} = 2k$.Let $k' = 2k = 2 	imes 1.0 = 2.0 	ext{ mol L}^{-1} 	ext{ min}^{-1}$.The integrated rate equation for this reaction is $[A]_t = [A]_0 - k't$.Given $[A]_0 = 2 	ext{ M}$.To complete $75\%$ of the reaction, the amount of $A$ consumed is $0.75 	imes [A]_0 = 0.75 	imes 2 = 1.5 	ext{ M}$.The remaining concentration $[A]_t = 2 - 1.5 = 0.5 	ext{ M}$.Substituting the values: $0.5 = 2 - 2.0 	imes t$.$2.0t = 1.5$.$t = \frac{1.5}{2.0} = 0.75 	ext{ min}$.

$Time \ (s)$	$Rate \ (mol \ L^{-1} s^{-1})$
$0$	$1.60 \times 10^{-2}$
$10$	$1.60 \times 10^{-2}$
$20$	$1.60 \times 10^{-2}$
$30$	$1.60 \times 10^{-2}$

$2A \xrightarrow{k} B$ is a zero-order reaction, where $k = 1.0 \text{ mol L}^{-1} \text{ min}^{-1}$. If the initial concentration of $A$ is $2 \text{ M}$, then the time taken to complete $75\%$ of the reaction will be (in $\text{ min}$)

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