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(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Given $t_{1/2} = 1 \ 	ext{hour} = 60 \ 	ext{min}$ and $[A]_0 =

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$15 \ 	ext{min}$

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(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Given $t_{1/2} = 1 \ 	ext{hour} = 60 \ 	ext{min}$ and $[A]_0 = 2.0 \ 	ext{mol L}^{-1}$.$60 \ 	ext{min} = \frac{2.0}{2k} \implies k = \frac{2.0}{2 	imes 60} = \frac{1}{60} \ 	ext{mol L}^{-1} \ 	ext{min}^{-1}$.For a zero order reaction,the integrated rate law is $[A]_t = [A]_0 - kt$.To find the time $t$ for concentration to decrease from $0.50 \ 	ext{mol L}^{-1}$ to $0.25 \ 	ext{mol L}^{-1}$:$t = \frac{[A]_0 - [A]_t}{k} = \frac{0.50 - 0.25}{1/60} = 0.25 	imes 60 = 15 \ 	ext{min}$.

Half life of zero order reaction $A \rightarrow \text{product}$ is $1 \ \text{hour}$,when initial concentration of reactant is $2.0 \ \text{mol L}^{-1}$. The time required to decrease concentration of $A$ from $0.50$ to $0.25 \ \text{mol L}^{-1}$ is:

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