For the reaction A rightarrow text{products}, | vedclass

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(A) From the graph,$t_{1/2} \propto [A]_0$,which indicates a zero-order reaction.For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2K}$.The slope of t

Vedclass · Accepted Answer

$2435$

Vedclass · Answer

(A) From the graph,$t_{1/2} \propto [A]_0$,which indicates a zero-order reaction.For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2K}$.The slope of the graph is $\frac{1}{2K} = 76.92 \ 	ext{min L mol}^{-1}$.Therefore,$K = \frac{1}{2 	imes 76.92} \ 	ext{mol L}^{-1} 	ext{min}^{-1} \approx 6.5 	imes 10^{-3} \ 	ext{mol L}^{-1} 	ext{min}^{-1}$.For a zero-order reaction,the concentration at time $t$ is given by $[A]_t = [A]_0 - Kt$.Given $[A]_0 = 2.5 \ 	ext{mol L}^{-1}$,$t = 10 \ 	ext{min}$,and $K = \frac{1}{153.84} \ 	ext{mol L}^{-1} 	ext{min}^{-1}$.$[A]_{10} = 2.5 - (\frac{1}{153.84}) 	imes 10 = 2.5 - 0.065 = 2.435 \ 	ext{mol L}^{-1}$.$[A]_{10} = 2435 	imes 10^{-3} \ 	ext{mol L}^{-1}$.

For the reaction $A \rightarrow \text{products}$,the graph of $t_{1/2}$ versus $[A]_0$ is given below. The concentration of $A$ at $10 \ \text{minutes}$ is $.......... \times 10^{-3} \ \text{mol L}^{-1}$ $(nearest \ integer)$. The reaction was started with $2.5 \ \text{mol L}^{-1}$ of $A$.

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