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(B) For a zero order reaction,the integrated rate equation is: $[A]_t = [A]_0 - kt$ Rearranging this gives: $[A]_0 - [A]_t = kt$ This represents a str

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$0.65$

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(B) For a zero order reaction,the integrated rate equation is: $[A]_t = [A]_0 - kt$ Rearranging this gives: $[A]_0 - [A]_t = kt$ This represents a straight line graph of $([A]_0 - [A]_t)$ versus $t$ with a slope equal to $k$. Given: Slope $(k)$ = $0.02 \ mol \ L^{-1} \ min^{-1}$ Time $(t)$ = $30 \ min$ Concentration at time $t$ $([A]_t)$ = $0.05 \ mol \ L^{-1}$ Using the equation $[A]_0 = [A]_t + kt$: $[A]_0 = 0.05 + (0.02 	imes 30)$ $[A]_0 = 0.05 + 0.60$ $[A]_0 = 0.65 \ mol \ L^{-1}$ Therefore,the initial concentration is $0.65 \ M$.

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