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(D) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Given $t_{1/2} = 1 \ h$ and $[A]_0 = 2.0 \ mol \ L^{-1}$,we have

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$0.25$

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(D) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Given $t_{1/2} = 1 \ h$ and $[A]_0 = 2.0 \ mol \ L^{-1}$,we have $1 = \frac{2.0}{2k}$,which gives the rate constant $k = 1.0 \ mol \ L^{-1} \ h^{-1}$.For a zero-order reaction,the rate law is $[A]_t = [A]_0 - kt$.To find the time $t$ for the concentration to change from $0.50 \ mol \ L^{-1}$ to $0.25 \ mol \ L^{-1}$,we use $\Delta [A] = k 	imes t$.$0.50 - 0.25 = 1.0 	imes t$.$0.25 = 1.0 	imes t$.Therefore,$t = 0.25 \ h$.

For a zero-order reaction $A \rightarrow \text{Product}$,the half-life is $1 \ h$. If the initial concentration of reactant $A$ is $2.0 \ mol \ L^{-1}$,how many hours will it take for the concentration to decrease from $0.50 \ mol \ L^{-1}$ to $0.25 \ mol \ L^{-1}$?

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