For a zero order reaction,a plot of t_{1/2} v | vedclass

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(D) For a zero order reaction:Rate $= K[A]^0 = K$Integrating the rate law,$[A]_t = [A]_0 - Kt$.At half-life $(t = t_{1/2})$,the concentration $[A]_t =

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$A$ straight line passing through the origin and slope $= 1/(2K)$

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(D) For a zero order reaction:Rate $= K[A]^0 = K$Integrating the rate law,$[A]_t = [A]_0 - Kt$.At half-life $(t = t_{1/2})$,the concentration $[A]_t = [A]_0/2$.Substituting these values: $[A]_0/2 = [A]_0 - K t_{1/2}$.Rearranging gives $K t_{1/2} = [A]_0/2$,or $t_{1/2} = \frac{[A]_0}{2K}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = t_{1/2}$,$x = [A]_0$,and $m = 1/(2K)$,the plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $1/(2K)$.

For a zero order reaction,a plot of $t_{1/2}$ versus $[A]_0$ will be

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