If c_o is the initial concentration of the re | vedclass

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(A) For a zero-order reaction,the integrated rate equation is given by: $[A] = [A]_0 - Kt$. At half-life $(t = t_{1/2})$,the concentration of the reac

Vedclass · Accepted Answer

$c_o / 2K$

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(A) For a zero-order reaction,the integrated rate equation is given by: $[A] = [A]_0 - Kt$. At half-life $(t = t_{1/2})$,the concentration of the reactant becomes half of its initial concentration,i.e.,$[A] = [A]_0 / 2$. Substituting these values into the equation: $[A]_0 / 2 = [A]_0 - Kt_{1/2}$. Rearranging for $t_{1/2}$: $Kt_{1/2} = [A]_0 - [A]_0 / 2 = [A]_0 / 2$. Therefore,$t_{1/2} = [A]_0 / 2K$. Given the initial concentration is $c_o$,the half-life is $c_o / 2K$.

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

If $c_o$ is the initial concentration of the reactant,what will be the half-life period for a zero-order reaction?

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